Decomposition of a polynomial of the fifth degree into quadratic factors using the Lagrange interpolation polynomial 1. Definition of the Lagrange interpolation polynomial of the fifth degree. To factor the reduced polynomial of the fifth degree into factors, it is necessary to satisfy the equality: f (x) = φ (x) g (x). Moreover, the degree of the polynomials φ (x) and g (x) should not be higher than five. To determine an integer polynomial not higher than the fifth degree with a given table of values, there is a formula for the Lagrange interpolation polynomial (IML): 6 Ak k = 1 F "(xk) (x − xk), where F (x) = (x-x1) · ( x-x2) (x-x3) (x-x4) (x- φ (x) = F (x) ∑ x5) (x-x6), Fʹ (xk) values of the derivative of the function F (x) at points xk. Where it is necessary to set the coordinates of six points on the plane. To determine the factors φ (x) and g (x), we arbitrarily choose six integer values x = x1; x2; x3; x4; x5; x6 and begin to substitute them into the equality f (x) = φ (x) g (x). We obtain: f (x1) = φ (x1) g (x1); f (x2) = φ (x2) g (x2); f (x3) = φ (x3) g (x3); f (x4) = φ (x4) g (x4); f (x5) = φ (x5) g (x5); f (x6) = φ (x6) G (x6). These equalities show that each value φ (xk) of the required factor φ (x) is a divisor of the number f (xk). To construct the factor φ (x), we use the IML and substitute arbitrary integers Аk, and we choose the values of xk in the form of consecutive integers close to zero, i.e. x1 = -3; x2 = -2; x3 = -1; x4 = 0; x5 = 1; x6 = 2. In this form, the IML φ (x) looks like this:
F (x) φ (x) A4 + A2 A3 + A1 A5 F "(1) (x − 1) + + A6 F" (- 3) (x + 3) F "(- 2) (x + 2) + + F "(0) x F" (- 1) (x + 1) F "(2) (x − 2)), · (where F (x) = (x + 3) · (x + 2) (X + 1) x (x-1) Definition: the numbers A1; A2; A3; A4; A5; A6 taken from the IML formula written in a series are called the Lagrangian series 2. Decomposition of the polynomial into linear factors using the IML Theorem 1 (Generalization of Horner's scheme) The polynomial φ (x) is linear if the numbers A1; A2; A3; A4; A5; A6 form an increasing sequence of integers Proof: we bring polynomial (2) to the lowest common denominator, i.e. to 120 F (x), the resulting numerator is written as a polynomial of the fifth degree in which the coefficients contain the numbers A1; A2; A3; A4; A5; A6. In order for the polynomial (2) to be linear, it is necessary to equate to zero the coefficients at "x" of the fifth, fourth, third and second degrees, and the coefficient at "X" of the first degree equate to 120. As a result So we get the following system of five equations with six variables: -A1 + 5A2-10A3 + 10A4-5A5 + A6 = 0 5A2-20A3 + 30A4-20A5 + 5 A6 = 0 5 A1-35 A2 + 70 A3-50 A4 + 5 A5 + 5 A6 = 0 -5 A2 + 80 A3-150 A4 + 80 A4-5 A6 = 0 -4 А1 + 30 А2-120 А3 + 40 А4 + 60 А5-6 А6 = 120. If you fix the number A6, then all the rest will be expressed by the following formulas: A1 = A6-5; A2 = A6-4; A3 = A6- 3; A4 = A6-2; A5 = A6-1.
We've got an ascending sequence of integers. The theorem implies that the linear factor has the following form: φ (x) = x + A4 (3). Definition: a sequence of numbers given by these ratios A1 = A6-5; A2 = A6-4; A3 = A6-3; A4 = A6-2; A5 = A6-1; A6 is called the linear Lagrangian series. Definition: a linear Lagrangian series is called a "candidate" if all its numbers Аk are divisors of the corresponding values of the function f (xk), where k = 1; 2; 3; 4; 5; 6. For all candidates, we construct a linear factor φ (x) according to formula (3) and check it for divisibility with f (x). It follows from the theorem that the linear factor has the following form φ (x) = x + A4, where A4 is the divisor of the free term, i.e. Similar to the given polynomial according to Horner's scheme. Example: f (x) = x5-8x4 + 2x3-16x2 + x-8. Using Horner's scheme, we find the value of the polynomial at x = -3; -2; -one; 0; 1; 2. To do this, let's compose Table 1: -8 -11 -10 -9 -8 -7 -6 -3 -2 -1 0 1 2 The last column of Table 1 will be rewritten as the first row of Table 2. Let's choose in this row the number with the smallest number of divisors. In our example, this number is -8. Let's write down all its divisors in a column. For each divisor of the number -8, write a linear Lagrangian series in a line. Let us choose "candidates" from the resulting Lagrangian series. Let us construct a polynomial φ (x) with respect to f (0) using the "candidates". linear factor -8 -1100 -250 -36 -8 -28 -150 is determined 1 1 1 1 1 1 1 2 35 22 11 2 -5 -10 -16 -121 -60 -27 -16 -21 -36 1 364 121 28 1 -20 -71
36 A3 0 -2 1 -3 3 -5 7 -9 -8 A4 1 -1 2 -2 4 -4 8 -8 -28 A5 2 0 3 -1 5 -3 9 -7 -150 A6 3 1 4 0 6 -2 10 -6 to formula (3) and check them for divisibility with the given polynomial f (x) = x5-8x4 + 2x3-16x2 + x-8. Table 2: -250 -1100 A2 A1 -2 -1 -3 -4 0 -1 -5 -4 2 1 -6 -7 5 6 -11 "candidate -10 at" which contain numbers that are not divisors of the corresponding values of the function f (x). This table contains a row or Lagrangian series of all numbers, which are divisors of the corresponding values of the function f (x). This row is the only candidate. In this series A4 = -8, substituting into the formula φ (x) = x- A4, we find φ (x) = x- 8. Highlight the real candidate in black. 3. Decomposition of the polynomial factors using the IML. Check: x5-8x4 + 2x3-16x2 + x-8 = (x-8) (x4 + 2x2 + 1). into quadratic Theorem 2. The factor φ (x) is quadratic if the numbers A1; A2; A3; A4; A5; A6 are interconnected by the following ratios: A1 = 5 (A5 + 4) -4 A6 A2 = 4 (A5 + 3) -3 A6 A3 = 3 (A5 + 2) -2 A6 A4 = 2 (A5 + 1) -1 A6
Proof: Proof: we bring polynomial (1) to the lowest common denominator, i.e. to 120 · F (x), the resulting numerator is written in the form of a polynomial of the fifth degree in which the coefficients contain the numbers A1; A2; A3; A4; A5; A6. In order for the polynomial (1) to be quadratic, it is necessary to equate to zero the coefficients at "x" of the fifth, fourth and third degrees, and the coefficient at "x" of the second degree equal to 120. As a result, we obtain the following system of four equations with six variables: -A1 + 5 A2-10 A3 + 10 A4-5 A5 + A6 = 0 5 A2-20 A3 + 30 A4-20 A5 + 5 A6 = 0 5 A1-35 A2 + 70 A3-50 A4 + 5 A5 + 5 A6 = 0 -5 A2 + 80 A3-150 A4 + 80 A5-5 A6 = 120. If we fix two numbers A5 and A6, then all the rest will be expressed by the following formulas: A1 = 5 · (A5 + 4) -4 · A6; A2 = 4 * (A5 + 3) -3 * A6; A3 = 3 * (A5 + 2) -2 * A6; A4 = 2 (A5 + 1) -1 A6. It follows from the theorem that the quadratic factor is expressed by the formula φ (x) = x2 + (A6- A5-3) x + A4. (4) Definition: A sequence of integers given by the following relationships; A3 = 3 * (A5 + 2) -2 * A6; А4 = 2 · (А5 + 1) -1 · А6 is called a quadratic Lagrangian series Definition: a quadratic Lagrangian series is called a "candidate" if all its numbers Аk are divisors of the corresponding values of the function f (xk), k = 1; 2; 3; 4 ; 5; 6. For all candidates, we construct a quadratic factor φ (x) according to formula (4) and check it for divisibility with f (x). A1 = 5 * (A5 + 4) -4 * A6; A2 = 4 (A5 + 3) -3 A6
А3 А4 + d + 4 А4 А5 + d + 2 А5 А5 4. Simplified form of quadratic Lagrangian series. The formulas for the quadratic Lagrangian series can be simplified. To do this, let us denote the difference A5-A6 by the letter "d", then the numbers of the quadratic Lagrangian series will look like simpler formulas and convenient for their construction: A1 A2 A2 + d + 8 A3 + d + 6 Example: A5 = 7; A6 = 10 compose a quadratic Lagrangian series. We find d = 7-10 = -3, then according to the formulas of the table we find the numbers of this series: A1 A2 + d + 8 10 + (- 3) +8 15 Answer: 15; 10; 7; 6; 7; 10. Consider an example of factoring a reduced polynomial of the fifth degree into factors: f (x) = x5-5x4 + 13x3-22x2 + 27x- 20. A5 A2 A3 + d + 6 A5 7 + (- 3) +6 6 + (- 3) +4 7 + (- 3) +2 7 7 10 A4 A5 + d + 2 A3 A4 + d + 4 7 6 A6 A6 A6 A6 10 10 1) Using Horner's scheme, we find the values of the function at x = -3; -2; -1; 0; 1; 2. To do this, make a table: 1 1 1 1 1 1 1 -5 -8 -7 -6 -5 -4 -3 13 37 27 19 13 9 7 -22 -133 -76 -41 -22 -13 -8 -3 - 2 -1 0 1 2 2) Determine whether the given polynomial has linear factors. To do this, write the resulting values of the function in line No. 3 of the table. From these we choose the number with the smallest number of divisors. In our example, this number is "2". Let's write down all its integer divisors in a column. For each divisor of the number "2" in -20 -1298 -378 -88 -20 -6 2 27 426 179 68 27 14 11
line we write down linear Lagrangian series. We select candidates from them and check for divisibility with the given polynomial f (x). Table 3: -1298 A1 -378 A2 -88 A3 -20 A4 -3 0 -4 -5 -6 A5 0 -2 1 -3 2 A6 1 -1 2 -2 which contain numbers that are not divisors of the corresponding values of the function f (x). There is no need to fill in empty cells, since the constructed Lagrangian quadratic series with a number in a gray cell is certainly not a "candidate". From this table No. 3 it is clear that there are no “candidates”. This means that the given polynomial f (x) = x5-5x4 + 13x3- 22x2 + 27x-20 cannot be decomposed into linear factors. 3) Determine whether a given polynomial has quadratic factors. To do this, write the resulting values of the function in line No. 4 of the table. Of these, we choose two numbers with the smallest number of divisors. In our example, these are the numbers "2" and "-6", we will write their divisors into columns. For each pair of divisors of the numbers "2" and "-6" we write down the quadratic Lagrangian series in a line. We select candidates from them and check them for divisibility with a given polynomial f (x). Table No. 4: -1298 A1 A2 + d + 8 -378 A2 A3 + d + 6 5 -88 A3 A4 + d + 4 1 10 -5 -20 A4 A5 + d + 2 3 -1 5 -3 7 -5 -6 A5 A5 1 -1 2 -2 3 -3 2 A6 A6 1 1 1 1 1 1 dd = A5- A6 d = 0 d = -2 d = 1 d = -3 d = 2 d = -4
19 7 2 14 -2 14 7 22 2 13 6 11 5 2 5 -1 8 -4 7 19 1 13 -11 5 1 7 -1 9 -3 15 -9 2 -2 4 -4 6 -6 12 -12 6 2 8 0 10 -2 16 -8 6 -6 1 -1 2 -2 3 -3 6 -6 1 -1 2 -2 3 -3 6 -6 1 -1 2 -2 3 -3 6 -6 1 1 -1 -1 -1 -1 -1 -1 -1 -1 2 2 2 2 2 2 2 -2 -2 -2 -2 -2 -2 -2 -2 d = 5 d = -7 d = 2 d = 0 d = 3 d = -1 d = 4 d = -2 d = 7 d = -5 d = -1 d = -3 d = 0 d = -4 d = 1 d = -5 d = 4 d = -8 d = 3 d = 1 d = 4 d = 0 d = 5 d = -1 d = 8 d = -4 "cand." "Cand." In this table number 4, we see two "candidates". With their help, according to the formula φ (x) = x2 + (A6- A5-3) x + A4, we find the square factors: φ1 (x) = x2-3x + 4; φ2 (x) = x2 + x-4. The check shows that one of the two factors is true, this is φ1 (x) = x2-3x + 4, and the other factor turned out to be extraneous. Answer: x5-5x4 + 13x3-22x2 + 27x-20 = (x2-3x + 4) (x3-2x2 + 3x-5). In this table number 4 received 32 quadratic Lagrangian series. This number is determined by the number of different pairs of divisors, both positive and negative, which are located in two columns next to each other. two function values,
5. Decrease in the number of quadratic Lagrangian series. By definition, If the values of the function, the number of divisors, which are minimal, are not located in the neighborhood, then you can use the following theorem: Theorem 3 Let A4 and A6 are known then A5 = (A4 + A6 1): 2-1 Let A3 and A6 are known then A5 = (A3 + A6 2): 3-2 Let A2 and A6 are known then A5 = (A2 + A6 3): 4-3 Let A1 and A6 are known then A5 = (A1 + A6 4): 5-4. Proof: Let us prove the last equality A5 = (A1 + A6 4): 5-4. quadratic Lagrangian numbers, A1 = 5 (A5 + 4) -4 A6 substitute this number in the original equality, we get A5 = (5 A5 + 20): 5-4 = A5 + 4-4 = A5 as required. Other equalities are proved in a similar way. This theorem makes it possible to reduce the number of quadratic Lagrangian series. Consider the example we have already solved f (x) = x5-5x4 + 13x3-22x2 + 27x-20 and solve it in the case when we consider quadratic Lagrangian series built using the divisors A4 and A6. Table No. 5: -1298 -378 A2 A1 A2 + A3 + d + 6 d + 8 dd = A5- A6 -88 A3 A4 + d + 4 -20 A4 A5 + d + 2 1 -1 5 -5 1 -1 -6 A5 ( A4 + A6 1): 2 -1 0 -1 2 -3 -1 -2 2 A 6 A 6 1 1 d = -2 1 d = 1 1 d = -4 - d = 0 1 d = -1 - 1 5 7 1 10 -5 5 2 14
19 11 7 22 2 2 14 -2 13 6 5 -1 8 -4 7 1 19 5 -5 2 -2 4 -4 10 -10 20 -20 2 -2 4 -4 10 -10 20 -20 1 -4 1 -1 2 -2 5 -5 10 -10 -1 -3 0 -4 3 -7 8 -12 "cand." "Cand." d = 2 - 1 - 1 2 d = -1 2 d = -3 2 d = 0 2 d = -4 2 2 2 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 d = 1 d = -1 d = 5 In this table number 5, we received 24 quadratic Lagrangian series. Since in the formula the sum of A4 and A6 must be divided by 2, therefore, the divisors A4 and A6 must be either both even or both odd. Due to this, the number of quadratic Lagrangian series has decreased. If we use this Theorem 3 to write quadratic Lagrangian series built using A1 and A6, then the number of series will decrease to 12. Table 6: -378 -1298 A1 A2 2 A6 d -88 A3 -20 A4 -6 A5
"Cand." A3 + d + 6 5 d = -4 d = 0 "cand." "Cand." A5 + d + 2 -5 -1 A4 + d + 4 -5 1 (4A1 + A6): 5-4 -3 -1 -15 -5 -7 7 -2 2 -26 -6 -10 12 A6 d = A5- A6 d = -4 1 1 d = -2 1 -1 -1 -1 2 2 2 -2 d = -4 -2 -2 A2 + d + 8 1 11 -59 -1 -11 -59 2 22 -118 - 2 -22 118 In table 6, the number of quadratic Lagrangian series has decreased to 12, since A5 is found by the formula (4A1 + A6): 5-4 and A5 as an integer must be less than or equal to -6. In all tables, the black highlighted line is a “valid candidate”. The rest of the candidates are "sham". For a polynomial of the sixth degree, one can prove that the quadratic factor can be found by the formula: φ (x) = x2 + (A7 - A6 - 5) · x + A4, where the numbers are A1; A2; A3; A4; A5; A6; A7 form a quadratic Lagrangian series. 6. Conclusions: 1. This method decomposition using IML -2 14 -4 8 -4 4 -8 is a generalization of the "Horner scheme". 2. This method can be used to determine quadratic factors for polynomials above the fifth degree. 3. This method can be used to investigate the properties of Lagrangian numbers for determining cubic polynomials in the decomposition of polynomials of the fifth and higher degree. 7. Literature: 1. A. N. Chebotarev "Foundations of Galois theory", OMTI GTTI, 1934, 1 hour.
2. "Numbers and polynomials", compiled by А.А. Egorov - M .: Bureau Quantum, 2000 / supplement to the magazine "Kvant" No. 6, 2000.
In order to factorize, it is necessary to simplify the expressions. This is necessary in order to be able to further reduce. The decomposition of a polynomial makes sense when its degree is at least two. A polynomial with the first degree is called linear.
The article will reveal all the concepts of decomposition, theoretical foundations and methods of factoring a polynomial into factors.
Theory
Theorem 1When any polynomial with degree n, having the form P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0, are represented as a product with a constant factor with the highest power an and n linear factors (x - xi), i = 1, 2, ..., n, then P n (x) = an (x - xn) (x - xn - 1) ... ... · (X - x 1), where x i, i = 1, 2,…, n - these are the roots of the polynomial.
The theorem is intended for roots of complex type x i, i = 1, 2,…, n and for complex coefficients a k, k = 0, 1, 2,…, n. This is the basis of any decomposition.
When coefficients of the form a k, k = 0, 1, 2, ..., n are real numbers, then complex roots that will meet in conjugate pairs. For example, the roots x 1 and x 2 referring to a polynomial of the form P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 are considered to be complex conjugate, then the other roots are real, hence we obtain that the polynomial takes the form P n (x) = a n (x - x n) (x - x n - 1) ·. ... ... (X - x 3) x 2 + p x + q, where x 2 + p x + q = (x - x 1) (x - x 2).
Comment
The roots of the polynomial can be repeated. Consider the proof of the algebra theorem, a corollary of Bezout's theorem.
The main theorem of algebra
Theorem 2Any polynomial with degree n has at least one root.
Bezout's theorem
After the division of a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + has been made. ... ... + a 1 x + a 0 on (x - s), then we get the remainder, which is equal to the polynomial at the point s, then we get
P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 = (x - s) Q n - 1 (x) + P n (s), where Q n - 1 (x) is a polynomial of degree n - 1.
Corollary from Bezout's theorem
When the root of the polynomial P n (x) is considered s, then P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 = (x - s) Q n - 1 (x). This corollary is sufficient when used to describe a solution.
Factoring a square trinomial
A square trinomial of the form a x 2 + b x + c can be decomposed into linear factors. then we get that a x 2 + b x + c = a (x - x 1) (x - x 2), where x 1 and x 2 are roots (complex or real).
From this it can be seen that the expansion itself is reduced to solving the quadratic equation later.
Example 1
Factorize a square trinomial.
Solution
Find the roots of the equation 4 x 2 - 5 x + 1 = 0. To do this, you need to find the value of the discriminant by the formula, then we get D = (- 5) 2 - 4 · 4 · 1 = 9. Hence we have that
x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1
From this we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.
The brackets must be expanded to perform the check. Then we get an expression of the form:
4 x - 1 4 x - 1 = 4 x 2 - x - 1 4 x + 1 4 = 4 x 2 - 5 x + 1
After checking, we come to the original expression. That is, we can conclude that the decomposition is correct.
Example 2
Factor a square trinomial of the form 3 x 2 - 7 x - 11.
Solution
We get that it is necessary to calculate the resulting quadratic equation of the form 3 x 2 - 7 x - 11 = 0.
To find the roots, you need to determine the value of the discriminant. We get that
3 x 2 - 7 x - 11 = 0 D = (- 7) 2 - 4 3 (- 11) = 181 x 1 = 7 + D 2 3 = 7 + 181 6 x 2 = 7 - D 2 3 = 7 - 181 6
From this we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6.
Example 3
Factor the polynomial 2 x 2 + 1.
Solution
Now you need to solve the quadratic equation 2 x 2 + 1 = 0 and find its roots. We get that
2 x 2 + 1 = 0 x 2 = - 1 2 x 1 = - 1 2 = 1 2 i x 2 = - 1 2 = - 1 2 i
These roots are called complex conjugate, which means that the decomposition itself can be represented as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.
Example 4
Decompose the square trinomial x 2 + 1 3 x + 1.
Solution
First you need to solve a quadratic equation of the form x 2 + 1 3 x + 1 = 0 and find its roots.
x 2 + 1 3 x + 1 = 0 D = 1 3 2 - 4 1 1 = - 35 9 x 1 = - 1 3 + D 2 1 = - 1 3 + 35 3 i 2 = - 1 + 35 i 6 = - 1 6 + 35 6 ix 2 = - 1 3 - D 2 1 = - 1 3 - 35 3 i 2 = - 1 - 35 i 6 = - 1 6 - 35 6 i
Having received the roots, we write
x 2 + 1 3 x + 1 = x - - 1 6 + 35 6 i x - - 1 6 - 35 6 i = = x + 1 6 - 35 6 i x + 1 6 + 35 6 i
Comment
If the value of the discriminant is negative, then the polynomials remain second-order polynomials. Hence it follows that we will not decompose them into linear factors.
Methods for factoring polynomials of degree higher than two
The decomposition assumes a universal method. Most of all cases are based on a corollary of Bezout's theorem. To do this, you need to select the value of the root x 1 and lower its degree by dividing by a polynomial by 1 by dividing by (x - x 1). The resulting polynomial needs to find the root x 2, and the search process is cyclical until we get a complete decomposition.
If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic assumes the solution of equations with higher degrees and integer coefficients.
Taking the common factor out of parentheses
Consider the case when the free term is equal to zero, then the form of the polynomial becomes P n (x) = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x.
It can be seen that the root of such a polynomial will be equal to x 1 = 0, then the polynomial can be represented as the expression P n (x) = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x = = x (a n x n - 1 + a n - 1 x n - 2 +... + a 1)
This method is considered as taking the common factor out of the parentheses.
Example 5
Factor the third degree polynomial 4 x 3 + 8 x 2 - x.
Solution
We see that x 1 = 0 is the root of the given polynomial, then we can take x outside the brackets of the entire expression. We get:
4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)
We turn to finding the roots of the square trinomial 4 x 2 + 8 x - 1. Let's find the discriminant and roots:
D = 8 2 - 4 4 (- 1) = 80 x 1 = - 8 + D 2 4 = - 1 + 5 2 x 2 = - 8 - D 2 4 = - 1 - 5 2
Then it follows that
4 x 3 + 8 x 2 - x = x 4 x 2 + 8 x - 1 = = 4 xx - - 1 + 5 2 x - - 1 - 5 2 = = 4 xx + 1 - 5 2 x + 1 + 5 2
To begin with, let us consider the decomposition method containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0, where the coefficient at the highest power is 1.
When a polynomial has integral roots, then they are considered divisors of the free term.
Example 6
Expand the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.
Solution
Consider if there are whole roots. It is necessary to write down the divisors of the number - 18. We get that ± 1, ± 2, ± 3, ± 6, ± 9, ± 18. It follows that this polynomial has integral roots. You can check the Horner scheme. It is very convenient and allows you to quickly obtain the coefficients of the expansion of the polynomial:
It follows that x = 2 and x = - 3 are the roots of the original polynomial, which can be represented as a product of the form:
f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x 3 + 5 x 2 + 9 x + 9) = = (x - 2) (x + 3) (x 2 + 2 x + 3)
We pass to the decomposition of a square trinomial of the form x 2 + 2 x + 3.
Since the discriminant is negative, it means that there are no real roots.
Answer: f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x + 3) (x 2 + 2 x + 3)
Comment
It is allowed to use a selection of a root and division of a polynomial by a polynomial instead of the Horner scheme. We proceed to consider the expansion of a polynomial containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0, the oldest of which is equal to one.
This case takes place for rational fractional fractions.
Example 7
Factor f (x) = 2 x 3 + 19 x 2 + 41 x + 15.
Solution
It is necessary to change the variable y = 2 x, you should go to a polynomial with coefficients equal to 1 at the highest degree. You need to start by multiplying the expression by 4. We get that
4 f (x) = 2 3 x 3 + 19 2 2 x 2 + 82 2 x + 60 = = y 3 + 19 y 2 + 82 y + 60 = g (y)
When the resulting function of the form g (y) = y 3 + 19 y 2 + 82 y + 60 has integer roots, then finding them among the divisors of the free term. The entry will take the form:
± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, ± 12, ± 15, ± 20, ± 30, ± 60
Let us proceed to calculating the function g (y) at these points in order to obtain zero as a result. We get that
g (1) = 1 3 + 19 1 2 + 82 1 + 60 = 162 g (- 1) = (- 1) 3 + 19 (- 1) 2 + 82 (- 1) + 60 = - 4 g (2) = 2 3 + 19 2 2 + 82 2 + 60 = 308 g (- 2) = (- 2) 3 + 19 (- 2) 2 + 82 (- 2) + 60 = - 36 g (3) = 3 3 + 19 3 2 + 82 3 + 60 = 504 g (- 3) = (- 3) 3 + 19 (- 3) 2 + 82 (- 3) + 60 = - 42 g (4) = 4 3 + 19 4 2 + 82 4 + 60 = 756 g (- 4) = (- 4) 3 + 19 (- 4) 2 + 82 (- 4) + 60 = - 28 g (5) = 5 3 + 19 5 2 + 82 5 + 60 = 1070 g (- 5) = (- 5) 3 + 19 (- 5) 2 + 82 (- 5) + 60
We get that y = - 5 is the root of an equation of the form y 3 + 19 y 2 + 82 y + 60, which means that x = y 2 = - 5 2 is the root of the original function.
Example 8
It is necessary to divide with a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.
Solution
Let's write and get:
2 x 3 + 19 x 2 + 41 x + 15 = x + 5 2 (2 x 2 + 14 x + 6) = = 2 x + 5 2 (x 2 + 7 x + 3)
Checking the divisors will take a lot of time, so it is more profitable to take the factorization of the resulting square trinomial of the form x 2 + 7 x + 3. Equating to zero and find the discriminant.
x 2 + 7 x + 3 = 0 D = 7 2 - 4 1 3 = 37 x 1 = - 7 + 37 2 x 2 = - 7 - 37 2 ⇒ x 2 + 7 x + 3 = x + 7 2 - 37 2 x + 7 2 + 37 2
Hence it follows that
2 x 3 + 19 x 2 + 41 x + 15 = 2 x + 5 2 x 2 + 7 x + 3 = = 2 x + 5 2 x + 7 2 - 37 2 x + 7 2 + 37 2
Artificial tricks for factoring a polynomial
Rational roots are not inherent in all polynomials. To do this, you need to use special methods for finding the multipliers. But not all polynomials can be expanded or represented as a product.
Grouping method
There are times when you can group the terms of a polynomial to find the common factor and place it outside the brackets.
Example 9
Factorize the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.
Solution
Because the coefficients are integers, then the roots, presumably, can also be integers. To check, take the values 1, - 1, 2 and - 2 in order to calculate the value of the polynomial at these points. We get that
1 4 + 4 1 3 - 1 2 - 8 1 - 2 = - 6 ≠ 0 (- 1) 4 + 4 (- 1) 3 - (- 1) 2 - 8 (- 1) - 2 = 2 ≠ 0 2 4 + 4 2 3 - 2 2 - 8 2 - 2 = 26 ≠ 0 (- 2) 4 + 4 (- 2) 3 - (- 2) 2 - 8 (- 2) - 2 = - 6 ≠ 0
From this it is clear that there are no roots, it is necessary to use a different method of decomposition and solution.
It is necessary to group:
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 4 + 4 x 3 - 2 x 2 + x 2 - 8 x - 2 = = (x 4 - 2 x 2) + (4 x 3 - 8 x) + x 2 - 2 = = x 2 (x 2 - 2) + 4 x (x 2 - 2) + x 2 - 2 = = (x 2 - 2) (x 2 + 4 x + 1)
After grouping the original polynomial, it is necessary to represent it as the product of two square trinomials. To do this, we need to do the factorization. we get that
x 2 - 2 = 0 x 2 = 2 x 1 = 2 x 2 = - 2 ⇒ x 2 - 2 = x - 2 x + 2 x 2 + 4 x + 1 = 0 D = 4 2 - 4 1 1 = 12 x 1 = - 4 - D 2 1 = - 2 - 3 x 2 = - 4 - D 2 1 = - 2 - 3 ⇒ x 2 + 4 x + 1 = x + 2 - 3 x + 2 + 3
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 2 - 2 x 2 + 4 x + 1 = = x - 2 x + 2 x + 2 - 3 x + 2 + 3
Comment
The simplicity of the grouping does not mean that it is easy enough to choose the terms. There is no definite solution, so it is necessary to use special theorems and rules.
Example 10
Factor the polynomial x 4 + 3 x 3 - x 2 - 4 x + 2.
Solution
The given polynomial has no integral roots. It is necessary to group the terms. We get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = = (x 4 + x 3) + (2 x 3 + 2 x 2) + (- 2 x 2 - 2 x) - x 2 - 2 x + 2 = = x 2 (x 2 + x) + 2 x (x 2 + x) - 2 (x 2 + x) - (x 2 + 2 x - 2) = = (x 2 + x) (x 2 + 2 x - 2) - (x 2 + 2 x - 2) = (x 2 + x - 1) (x 2 + 2 x - 2)
After factoring, we get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = x 2 + x - 1 x 2 + 2 x - 2 = = x + 1 + 3 x + 1 - 3 x + 1 2 + 5 2 x + 1 2 - 5 2
Using abbreviated multiplication formulas and Newton's binomial for factoring a polynomial
The appearance often does not always make it clear which method should be used during decomposition. After the transformations have been made, you can build a line consisting of Pascal's triangle, otherwise they are called the Newton binomial.
Example 11
Factor the polynomial x 4 + 4 x 3 + 6 x 2 + 4 x - 2.
Solution
It is necessary to convert the expression to the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3
The expression x + 1 4 indicates the sequence of sum coefficients in parentheses.
Hence, we have x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3.
After applying the difference of the squares, we get
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3
Consider the expression in the second parenthesis. It is clear that there are no horses there, so the formula for the difference of squares should be applied again. We get an expression of the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3 = = x + 1 - 3 4 x + 1 + 3 4 x 2 + 2 x + 1 + 3
Example 12
Factor x 3 + 6 x 2 + 12 x + 6.
Solution
Let's do the transformation of the expression. We get that
x 3 + 6 x 2 + 12 x + 6 = x 3 + 3 2 x 2 + 3 2 2 x + 2 3 - 2 = (x + 2) 3 - 2
It is necessary to apply the formula for abbreviated multiplication of the difference of cubes. We get:
x 3 + 6 x 2 + 12 x + 6 = = (x + 2) 3 - 2 = = x + 2 - 2 3 x + 2 2 + 2 3 x + 2 + 4 3 = = x + 2 - 2 3 x 2 + x 2 + 2 3 + 4 + 2 2 3 + 4 3
A way to replace a variable when factoring a polynomial into factors
When changing a variable, the degree is reduced and the polynomial is decomposed into factors.
Example 13
Factor a polynomial of the form x 6 + 5 x 3 + 6.
Solution
According to the condition, it is clear that it is necessary to make the replacement y = x 3. We get:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6
The roots of the resulting quadratic equation are equal to y = - 2 and y = - 3, then
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3
It is necessary to apply the formula for abbreviated multiplication of the sum of cubes. We get expressions of the form:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3 = = x + 2 3 x 2 - 2 3 x + 4 3 x + 3 3 x 2 - 3 3 x + 9 3
That is, we got the required decomposition.
The cases discussed above will help in considering and factoring a polynomial in different ways.
If you notice an error in the text, please select it and press Ctrl + Enter
Keywords: equations, Polynomial, Equation roots
Lesson presentation
Back forward
Attention! Slide previews are for informational purposes only and may not represent all the presentation options. If you are interested in this work, please download the full version.
Lesson type: Lesson assimilation and consolidation of primary knowledge.
The purpose of the lesson:
- To acquaint students with the concept of roots of a polynomial, to teach how to find them. Improve the skills of applying Horner's scheme for expanding a polynomial in powers and dividing a polynomial by a binomial.
- Learn to find the roots of an equation using Horner's scheme.
- Develop abstract thinking.
- Foster a computing culture.
- Development of intersubject connections.
During the classes
1. Organizational moment.
Report the topic of the lesson, formulate goals.
2. Checking homework.
3. Learning new material.
Let F n (x) = a n x n + a n-1 x n-1 + ... + a 1 x + a 0 - polynomial with respect to x of degree n, where a 0, a 1, ..., an are given numbers, and a 0 is not equal to 0. If the polynomial F n (x) is divided with the remainder by the binomial xa, then the quotient (incomplete quotient) is polynomial Q n-1 (x) of degree n-1, the remainder R is a number, and the equality F n (x) = (x-a) Q n-1 (x) + R. The polynomial F n (x) is divisible entirely by the binomial (x-a) only in the case R = 0.
Bezout's theorem: The remainder R of dividing the polynomial F n (x) by the binomial (x-a) is equal to the value of the polynomial F n (x) for x = a, i.e. R = P n (a).
A bit of history. Bezout's theorem, despite its apparent simplicity and obviousness, is one of the fundamental theorems of the theory of polynomials. In this theorem, the algebraic properties of polynomials (which allow us to work with polynomials as with integers) are associated with their functional properties (which allow us to consider polynomials as functions). One of the ways to solve equations of higher degrees is to factorize the polynomial on the left side of the equation. The calculation of the coefficients of the polynomial and the remainder is written in the form of a table, which is called the Horner scheme.
Horner's scheme is an algorithm for dividing polynomials, written for the special case when the quotient is equal to the binomial x – a.
Horner William George (1786 - 1837), English mathematician. Major research relates to theory algebraic equations... Developed a method for the approximate solution of equations of any degree. In 1819 he introduced an important method for algebra for dividing a polynomial into binomials x - a (Horner's scheme).
Derivation of the general formula for Horner's scheme.
Dividing the polynomial f (x) into a binomial (x-c) with the remainder means to find a polynomial q (x) and a number r such that f (x) = (x-c) q (x) + r
Let's write this equality in detail:
f 0 xn + f 1 x n-1 + f 2 x n-2 + ... + f n-1 x + fn = (xc) (q 0 x n-1 + q 1 x n-2 + q 2 x n-3 + ... + q n-2 x + q n-1) + r
Let's equate the coefficients at the same degrees:
x n: f 0 = q 0 => q 0 = f 0 x n-1: f 1 = q 1 - c q 0 => q 1 = f 1 + c q 0 x n-2: f 2 = q 2 - c q 1 => q 2 = f 2 + c q 1 ... ... x 0: f n = q n - c q n-1 => q n = f n + c q n-1.
Demonstration of Horner's scheme by example.
Exercise 1. Using Horner's scheme, we divide with the remainder the polynomial f (x) = x 3 - 5x 2 + 8 by the binomial x-2.
| 1 | -5 | 0 | 8 | |
| 2 | 1 | 2*1+(-5)=-3 | 2*(-3)+0=-6 | 2*(-6)+8=-4 |
f (x) = x 3 - 5x 2 + 8 = (x-2) (x 2 -3x-6) -4, where g (x) = (x 2 -3x-6), r = -4 remainder.
Decomposition of a polynomial in powers of a binomial.
Using Horner's scheme, we expand the polynomial f (x) = x 3 + 3x 2 -2x + 4 in powers of the binomial (x + 2).
As a result, we should get the decomposition f (x) = x 3 + 3x 2 -2x + 4 = (x + 2) (x 2 + x-4) +12 = (x + 2) ((x-1) (x + 2) -2) +12 = (((1 * (x + 2) -3) (x + 2) -2) (x + 2)) + 12 = (x + 2) 3 -3 (x + 2 ) 2 -2 (x + 2) +12
Horner's scheme is often used to solve equations of the third, fourth, and higher degrees, when it is convenient to decompose a polynomial into a binomial x-a. Number a are called by the root of the polynomial F n (x) = f 0 x n + f 1 x n-1 + f 2 x n-2 + ... + f n-1 x + f n, if for x = a the value of the polynomial F n (x) is equal to zero: F n (a) = 0, i.e. if the polynomial is divisible entirely by the binomial x-a.
For example, the number 2 is the root of the polynomial F 3 (x) = 3x 3 -2x-20, since F 3 (2) = 0. it means. That the factorization of this polynomial contains the factor x-2.
F 3 (x) = 3x 3 -2x-20 = (x-2) (3x 2 + 6x + 10).
Any polynomial F n (x) of degree n 1 can have at most n real roots.
Any integer root of an equation with integer coefficients is a divisor of its intercept.
If the leading coefficient of the equation is 1, then all rational roots of the equation, if they exist, are integer.
Consolidation of the studied material.
To consolidate the new material, students are invited to complete the numbers from the textbooks 2.41 and 2.42 (p. 65).
(2 students solve at the blackboard, and the rest, having decided, in the notebook, check the tasks with the answers on the blackboard).
Summarizing.
Having understood the structure and principle of operation of Horner's scheme, it can also be used in computer science lessons, when the question of converting integers from the decimal number system to the binary one and vice versa is considered. The translation from one number system to another is based on the following general theorem
Theorem. To translate an integer Ap from p-ary numeral system to base numeral system d necessary Ap sequentially divide with remainder by a number d recorded in the same p-type system, until the obtained quotient becomes equal to zero. In this case, the remainder of the division will be d-similar digits of the number Ad, starting from the least significant bit to the highest. All actions must be carried out in p-ary numeral system. For a person, this rule is convenient only when p= 10, i.e. when translating from decimal system. As for the computer, on the contrary, it is “more convenient” for it to perform calculations in the binary system. Therefore, to translate “2 to 10”, successive division by ten in the binary system is used, and “10 to 2” is the addition of powers of ten. To optimize the calculations of the 10-in-2 procedure, the computer uses Horner's economical computational scheme.
Homework. It is proposed to complete two tasks.
1st. Using Horner's scheme, divide the polynomial f (x) = 2x 5 -x 4 -3x 3 + x-3 into the binomial (x-3).
2nd. Find the integer roots of the polynomial f (x) = x 4 -2x 3 + 2x 2 -x-6. (Considering that any integer root of an equation with integer coefficients is a divisor of its intercept)
Literature.
- Kurosh A.G. "Course of Higher Algebra".
- Nikolsky S.M., Potapov M.K. and others. Grade 10 “Algebra and the beginning of mathematical analysis”.
- http://inf.1september.ru/article.php?ID=200600907.
A task 1. Find the gcd of polynomials
f(x)=x 4 –2x 3 –x+2, g(x)=x 4 –x 3 +x–1, h(x)=x 4 –4x 2 –x+2.
Solution. The GCD of polynomials is found uniquely only up to a constant factor (constant nonzero factors do not affect the divisibility of polynomials). Therefore, we can agree to take the one with the leading coefficient equal to 1 as the GCD of polynomials.
Applying the Euclidean algorithm to polynomials with integer coefficients, we can, in order to avoid fractional coefficients, multiply the dividend or divisor by any non-zero number, and not only starting with any of the consecutive divisions, but also in the process of this division itself. This will lead, of course, to a distortion of the quotient, but the residuals of interest to us will acquire only a certain factor of zero degree.
To find the GCD of three polynomials, we first find the GCD of any two polynomials using the Euclidean algorithm, for example d(x)=(f(x),h(x)), and then find the GCD d(x) and g(x).
Euclid's algorithm consists in sequential division of polynomials with remainder. Let's divide first f(x) on the h(x), then h(x) by the remainder obtained by division r(NS) (first remainder), then the first remainder to the second remainder, etc., until we get zero at the remainder. GCD of polynomials f(x) and h(x) will be the last nonzero remainder. The division process will be carried out by "angle".
| _ | x 4 -2x 3 -x + 2 | x 4 -4x 2 -x + 2 | _ | x 4 -4x 2 -x + 2 | x 3 -2x 2 | ||||
| x 4 -4x 2 -x + 2 | 1 | x 4 -2x 3 | x + 2 | ||||||
| -2x 3 + 4x 2 | _ | 2x 3 -4x 2 -x + 2 | |||||||
| x 3 -2x 2 | 2x 3 -4x 2 | ||||||||
| _ | -x + 2 | ||||||||
| x-2 | |||||||||
| 0 | |||||||||
| _ | x 3 -2x 2 | x-2 | ||
| x 3 -2x 2 | x 2 | |||
| 0 | ||||
Hence the gcd of polynomials f(x) and h(x) is equal to the binomial x–2.
d(x)=(f(x), h(x))=x–2.
Similarly, we find the GCD of the polynomials d(x) and g(x), it will be equal to 1. Thus, ( f(x), g(x), h(x))=(g(x), (f(x), h(x)))=1.
Note . Sign "=" or "!!" means that in the course of division, a multiplication was performed by some number other than zero.
A task 2.Using Euclid's algorithm find the polynomials u(x) and v(x) satisfying the equality f(x)u(x)+g(x)v(x)=d(x), where d(x) - GCD of polynomials f(x) and g(x): f(x)=4x 4 –2x 3 –16x 2 +5x+9, g(x)=2x 3 –x 2 –5x+4.
Solution. Apply to polynomials f(x) and g(x) Euclid's algorithm. It must be remembered that here arbitrariness, which consisted in multiplying polynomials by constant factors, which is possible when finding the GCD, cannot be allowed, since here we will also use quotients, which can be distorted with the indicated arbitrariness.
As a result of division, we get:
f(x)=g(x)q 1 (x)+r 1 (x),
where q 1 (x)=2x, r 1 (x)= –6x 2 –3x+9,
g(x)=r 1 (x)q 2 (x)+r 2 (x),
where q 2 (x)= –x/3+1/3, r 2 (x)= –x+1,
r 1 (x)=r 2 (x)q 3 (x)+r 3 (x),
where q 3 (x)=6x+9, r 3 (x)=0.
Thus, Euclid's algorithm was written here in three lines, and the greatest common divisor is - r 2 (x)=x–1=d(x). To express d(x) in terms of polynomials f(x) and g(x), we find r 2 (x) from the second line of Euclid's algorithm:
r 2 (x)=g(x)–r 1 (x)q 2 (x).
Substituting into this equality instead of r 1 (x) its expression, found from the first line of the Euclidean algorithm, we get:
r 2 (x)=f(x)[–q 2 (x)]+g(x),
to get equality f(x)u(x)+g(x)v(x)=d(x), you need to multiply the previous equality by (–1), we get:
–r 2 (x)=f(x)q 2 (x) +g(x)[–1–q 1 (x)q 2 (x)]=d(x),
where u(x)=q 2 (x), v(x)= –1–q 1 (x)q 2 (x).
After substituting the polynomials into this equality q 1 (x), q 2 (x) we get:
u(x)= , v(x)= .
A task 3. Using the method of undefined coefficients, choose the polynomials u(x) and v(x) so that f(x)u(x)+g(x)v(x) = 1, (1) for polynomials f(x)=x 2 –2x–1, g(x)=2x 4 –3x 3 –6x 2 +2x+2.
Solution. Let us use the theorem: if d(x) is the gcd of polynomials f(x) and g(x), then one can find such polynomials u(x) and v(x), what
f(x)u(x)+g(x)v(x)=d(x).
In this case, it can be considered if the degrees of the polynomials f(x) and g(x) is greater than zero, so that the degree u(x) less degree g(x), and the degree v(x) less degree f(x).
Polynomials f(x) and g(x) satisfy equality (1) if ( f(x),g(x)) = 1. In our case f(x) and g(x) are coprime polynomials, which means that one can find the polynomial u(x)=ax 3 +bx 2 +cx+d and polynomial v(x)=ex+f.
Substituting into equality (1) instead of f(x), g(x), u(x), v(x) their expressions, we get:
(x 2 – 2x– 1)(ax 3 + bx 2 + cx + d)+(2x 4 – 3x 3 – 6x 2 + 2x + 2)(ex + f)=1
(a + 2e)x 5 + (b– 2a + 2f– 3e)x 4 + (c– 2b – a– 3f– 6e)x 3 + (d– 2c – b– 6f + 2e)x 2 +(–2d – c + 2f + 2e)x –– d + 2f = 1.
Thus, we have the equality of two polynomials: on the left-hand side there is a fifth-degree polynomial with undefined coefficients, and on the right-hand side there is a zero-degree polynomial. Two polynomials are equal if their coefficients are equal for the same powers of the unknown.
Equating the coefficients at the same powers of the unknown, we obtain a system of six linear equations with unknown a, b, c, d, e, f:
Solving it, we get: d = 3, e =–1, f = 2, c =–4, b =–3, a = 2.
Thus, the required polynomials u(x) and v(x) will be:
u(x)=2x 3 –3x 2 –4x+3, v(x)= –x+2.
A task 4. Using Horner's scheme, calculate f(but) and expand the polynomial f(x) in powers x–but, where f(x)=x 4 +2x 3 –7x 2 +3NS–1, but=2.
Solution. By Bezout's theorem, the remainder of the division of the polynomial f(x) to a linear binomial x–but equal to the value f(but) of the polynomial for x=but.
The division by "angle" can be written more simply: if f(x)=a 0 x n+a 1 x n –1 +a 2 x n– 2 + …+a n –1 x+a n, then the coefficients of the quotient q(x)=b 0 x n – 1 + b 1 x n –2 + b 2 x n –3 + …+b n-1 and remainder r from division f(x) on the x–a can be found by Horner's scheme:
f(2)=9=r 1, and the quotient of division f(x) on the x–2 there is q 1 (x)=x 3 +4x 2 +x+5, i.e. f(x)=
=(x–2)q 1 (x)+r 1
Then, according to Horner's scheme, we divide q 1 (x) on the x–2, we get the quotient q 2 (x) and the remainder r 2, then q 2 (x) we divide by x–2, we get q 3 (x) and r 3, etc.
For polynomial f(x) we get:
f(x)=(x–2)q 1 (x)+r 1 =(x–2)[(x–2)q 2 (x)+r 2 ]+r 1 =(x–2) 2 q 2 (x)+r 2 (x–2)+r 1 =
=(x––2) 2 [(x–2)q 3 (x)+r 3 ]+r 2 (x–2)+r 1 =(x–2) 3 q 3 (x)+r 3 (x–2) 2 +r 2 (x–2)+r 1 =
=(x–2) 3 [(x––2)q 4 (x)+r 4 ]+r 3 (x–2) 2 +r 2 (x–2)+r 1 =(x–2) 4 q 4 (x)+r 4 (x–2) 3 +r 3 (x–2) 2 +r 2 (x–2)+ +r 1 = r 5 (x–2) 4 +r 4 (x–2) 3 +r 3 (x–2) 2 +r 2 (x–2)+r 1.
Thus, the coefficients in the expansion of the polynomial f(x) in powers x–2 are equal respectively to the remainder of the division of the polynomials f(x), q 1 (x), q 2 (x), q 3 (x), q 4 (x) on the x–2.
The entire solution can be written into a table:
| –7 | –1 | ||||
The table shows that r 5 =1, r 4 =10, r 3 =29, r 2 =31, r 1 = 9 and
f(x)= (x–2) 4 +10(x–2) 3 +29(x–2) 2 +31(x–2)+9.
A task 5. Prove that.
Solution. Consider a polynomial. Number NS= –1 is the root of the polynomial f(x) and by Bezout's theorem f(x) is divisible by NS+1, i.e. f(x)=(x+1)g(x), where g(x) Is a polynomial with integer coefficients, therefore NS 11 +1 is divided into NS+1 for any integer NS... We put NS= 3 5. We get, i.e. , and since , we conclude that.
Comment... From the rules of "dividing by an angle" of a polynomial f(x) by the polynomial g(x) it is directly seen that if the polynomials f(x) and g(x) with integer coefficients, and g(x) reduced, then the quotient and remainder are polynomials with integer coefficients.
A task 6. Remainders of a polynomial division f(x) into binomials NS+5 and NS-3 are equal to -9 and 7, respectively. Find the remainders of dividing this polynomial by a polynomial g(x)=(x+5)(x-3).
Solution. By Bezout's theorem f(–5)= –9, f(3) = 7. When dividing a polynomial f(x) by the polynomial g(x)=x 2 +2x–15 we get some quotient q(x) and the remainder p(x)=ax+b, i.e. f(x)=(x 2 +2x–15)q(x)+(ax+b) .
Substituting in the last equality instead of NS values –5 and 3 we obtain a system of two equations with two unknowns a and b:
Having solved it, we find a=2, b= 1. Then the required remainder of the division of the polynomial f(x) by the polynomial g(x) will be equal to 2 NS+1.
A task 7. Given a polynomial f(x) with integer coefficients and. Prove that.
Solution. Consider the decomposition of the polynomial f(x) in powers ( x–10):
in view of the fact that it is divisible by 21, i.e. is divisible by 7. Similarly divisible by 3. Due to the mutual simplicity of 3 and 7, the number f(10)=a n is a multiple of 21.
A task 8. Expand the polynomial x 7 +3 to the product of polynomials of at most second degree with real coefficients.
Solution. Find the roots of the polynomial x 7 +3, they will be
Giving k values 0, 1, ..., 6, we get seven roots of the polynomial x 7 +3;
x 0 = ; x 1 = ; x 2 = ;
x 3 = = – ; x 4 = = ;
x 5 = = ;
x 6 = = .
Among them, only one valid is x 3 = -, the rest are complex, and pairwise conjugate: x 6 = , x 5 = , x 4 =. In general
X k =, x k= .
Consider the product
(x–x k)(x– )=(x 2 –(x k+ )x+x k)=x 2 – x+ where k=0, 1, 2.
We have a square trinomial with real coefficients. Polynomial x 7 +3 can be expanded into the product of 7 linear factors (a consequence of the main theorem of algebra). Multiplying the factors that correspond to the conjugate roots, we get the required decomposition:
x 7 +3=(x–x 0)(x–x 1)(x–x 2)(x–x 3)(x–x 4)(x–x 5)(x–x 6)=(x–x 3)(x–x 0)(x–x 6)(x–x 1)
(x–x 5)(x–x 2)(x––X 4)=(x–x 3)(x–x 0)(x– )(x–x 1)(x– )(x–x 2)(x– )=(x+ )
(x 2 - (2) x+ )(x 2 - (2) x+ ) (x 2 –– (2 ·) x+ ).
A task 9. Present the polynomial as the sum of the squares of two polynomials.
Solution. Any polynomial f(x) with real coefficients, positive for any is represented as the sum of squares of two polynomials. To do this, we find the roots of the polynomial f(x):, we decompose into linear factors, then we multiply and, we get the required representation:
We denote,, we get f(x)=p 2 (x)+q 2 (x).
A task 10. Determine the multiplicity of the root of the polynomial. Find the polynomial of greatest degree with simple roots, each root of which is a root of the polynomial f(x).
1) Check if the root of the polynomial f(x).
2) Check if the root of the first derivative of the polynomial f(x)
. f¢ (–1) = 0, therefore is the root
polynomial f(x), multiplicity at least 2.
3), so the root of multiplicity is at least 3.
4), the root of the polynomial f(x) of multiplicity 3, i.e. ... To find the polynomial of greatest degree with simple roots, each root of which is a root f(x), you need in the polynomial f(x) get rid of multiple roots. To do this, we divide the polynomial f(x) by the greatest common divisor of polynomials f(x) and f¢( x):. Therefore, the required polynomial will be, where, NS= 2 - simple roots of the polynomial.
Note: The multiplicity of the root could be checked according to Horner's scheme.
A task 11. Separate multiple factors of the polynomial
Solution. By the multiple factor theorem: if some irreducible polynomial over the field P g(x) is an k- multiple of the polynomial f(x) with coefficients from the field P, then g(x) is an ( k–1) - a multiple factor of the derivative f(x). Thus, when passing from f(x) To f′( x) the multiplicity of all factors decreases by 1. However, the polynomial f′( x) there may be factors that are not present in f(x). To get rid of them we will find GCD f(x) and f′( x). It will include only those factors that are included in f(x), but with 1 fold less.
Applying Euclid's algorithm, we get
Since there is a polynomial of the third degree, the factorization of which is generally difficult in the general case, but which, in turn, can have multiple factors, we will apply to it a similar process of decreasing the multiplicity of factors. We will receive. So the multiplier NS–1 enters with multiplicity 1, and therefore, it enters with multiplicity 2. Divide by ( NS–1) 2, we find. Hence we have: the factor ( NS–1) is included in f(x) with multiplicity 3, and NS+3 with a multiplicity of 2. Dividing f(x) by a polynomial, we get
A task 12. Prove that the number is irrational.
Solution. This number is the root of the reduced integer polynomial, which has no rational roots, since all its rational roots are integers and must be divisors of 5.
A task 13. Find rational roots of a polynomial
f(x)=6x 4 +19x 3 –7x 2 –26x+12.
Solution. If a rational irreducible fraction that is a root of a polynomial f(x)=but 0 x n + a 1 x n– 1 + a 2 x n– 2 +… + A n– 1 x + a n with integer coefficients, then:
1. k there is a divisor but 0 ;
2. p there is a divisor a n;
3. p – mk there is a divisor f(m) for any integer m.
In our case: k can take values: ± 1, ± 2, ± 3, ± 6, and p- ± 1, ± 2, ± 3, ± 4, ± 6, ± 12. Now it would be possible to check each of these numbers of the form by substitution in a polynomial or according to Horner's scheme. However, many of these numbers can be weeded out more simple way... Let us find the boundaries of the real roots of the given polynomial VG x = 1 +, NG x = - (1+), where BUT Is the largest of the absolute values of the coefficients, and but 0 - coefficient at x n or VG x = 1 +, where k- the index of the first negative coefficient of the polynomial f(x), but B- the largest of the absolute values of its negative coefficients (this method is applicable when but 0> 0). In our example k=2, B=26, but 0 = 6. VG x = 1 +< 4.
To find the lower boundary in this way, it is sufficient in f(x) instead of x substitute (- x) and use the following rule: the lower bound of the real roots of the polynomial f(x) is equal to the upper bound of the real roots of the polynomial f(–x) taken with the opposite sign. In our case
f(–x)=6x 4 –19x 3 –7x 2 +26x+12, and 0 = 6, k=1, B= 19. VG x = 1 +<5, значит, нижняя граница – НГ х = –5. Итак, корни многочлена заключены в интервале (–5,4). Более точные границы можно было найти по методу Ньютона. Воспользуемся еще тем, что если – корень f(x), then the whole. Find f(1)=4,
f(–1) = 13, so - whole, - whole, if - root f(x).
We check all possible fractions, taking into account the boundaries of the roots.
| c | d | c | c | d | d | c | d | c | d | c | d | c | d | c | c | d | d | |
| c | d | c | d | d | d | c | d | c |
In the course of such a check, rational numbers 2, –3,, - "candidates for roots" appeared, we check them according to Horner's scheme, we make sure that f(2)≠0, , f(–3) = 0,. For a polynomial of the fourth degree, two roots were found, which means that f(x) multiple ( x+3) or f(x)=(6x 2 +4x–8)(x+3). Polynomial roots g(x)=6x 2 +4x–8 we find directly x= - irrational numbers.
A task 14. Prove that this equation has no nonzero integer solutions.
Solution. The left-hand side of the equality is a homogeneous fourth degree polynomial. We divide both sides of the equality by NS 4 . We get
Let's put, then. A given equation has a nonzero integer solution if and only if the polynomial has rational roots. Reduced polynomial, integral, all its rational roots are: first, whole; secondly, by the divisors of the free term 9, i.e. must belong to the set (± 1, ± 3, ± 9). By direct verification, one can make sure that not a single element of the given set is a root of the polynomial, i.e. this polynomial has no rational roots, which means that the given equation has nonzero integer roots.
A task 15. Under what natural n will be a prime number?
Solution. Let us show that. Indeed, if but Is an arbitrary root of the polynomial, then but will be the root of the polynomial, i.e. but 3 = 1 and but 2 +but+1=0.
Consider, i.e. but Is the root of the polynomial. As but Is an arbitrary root of a polynomial, then each root of a polynomial is a root of a polynomial, therefore, where P(x) Is a polynomial with integer coefficients.
Suppose then, i.e. ...
Consider the cases and.
2. When is a prime number.
A natural number is represented as a product of two natural numbers. This shows what can be simple if or, - we discard.
At, and is presented as a product of two natural numbers exceeding 1, which means that this number is composite.
A task 16. Solve the equations in the field of complex numbers:
1)x 3 +6x+2=0; 2) x 3 –9x 2 +18x–28=0; 3) x 4 -2x 3 +4x 2 -2x + 3=0.
1. Let's solve the equation x 3 +6x+2=0.
For the roots of the cubic equation x 3 +ax+b= 0 there is the so-called Cardano formula: x i = u i + v i (i= 0, 1, 2), where u 0 , u 1 , u 2 - the value of the radical
u= and v i=. In our case, but=6, b=2,
u= = = = = (cos + i sin), where l= 0, 1, 2. Substituting instead of l values 0, 1, 2, we get: u 0 = , u 1 =
= (cos + i sin) = (- + i), u 2 = (cos + i sin) = (- - i ),
v 0 = = = = ,
v 1 = = = = ( +i ),
v 2 = = = = ( –i ),
x 0 = u 0 +v 0 = – , x 1 =u 1 +v 1 = , x 2 = u 2 +v 2 =
Answer: - ; ...
2. Solve the equation x 3 –9x 2 +18x–28=0.
Let us reduce our equation to an equation of the form y 3 +ay+b= 0 by substituting x=y– =y+3, (a 0 , a 1 - coefficients at x 3 and x 2). We get:
y 3 –9y–28 = 0. His solutions are found using the Cardano formula: y i = u i+ v i, (i=0, 1,…2),
where u 0 =3, u 1 = , u 2 = , v 0 =1 , v 1 = , v 2= ,
y 0 =4, y 1 = , y 2 = , x 0 =7, x 1 = , x 2 = .
Answer: 7; ...
3. Solve the equation x 4 -2x 3 +4x 2 -2x + 3=0.
Let's apply the Ferrari way. We leave on the left side of the equation the terms with NS 4 and NS 3 and complement it to a full square:
Now we add to both parts the terms with the new unknown y so that the left side becomes a square again (regardless of the value y)
Here the coefficients in front of the powers x on the right side depend on the undefined value y... Let's choose the value of y so that the right side becomes a square. For this, it is necessary that the discriminant of the square (with respect to x) of the trinomial on the right-hand side was equal to zero. Equating this discriminant to zero, we get:
from here y= 4 and.
Substituting y= 4 into the equation (*), we get: or. Taking the square root from both sides of the resulting equation, we get two quadratic equations: and or and. Having solved them, we will find 4 roots of our equation:,.
Answer: , .
A task 17. Given polynomials
f(x)=x 3 –3x 2 +2x–5, g(x)=x 3 +3x 2 –1.
1) Determine the number of real roots of each;
2) Using Sturm's theorem, find the interval ( a, b), where b – a= 1 containing the largest root x 0 polynomial g(x);
3) Calculate with an accuracy of 0.0001 root x 0 using the linear interpolation method and Newton's method;
1. If the coefficients a and b equations x 3 +ax+b= 0 are real, then the number of real roots of this equation is completely determined by the sign of the number D = – 4a 3 – 27b 2, called the discriminant of the polynomial x 3 +ax+b, in the following way:
a) for D = 0, all three roots are real, of which two are equal;
b) for D> 0 - all three roots are real;
c) for D<0 – один корень действительный, два мнимых.
In our case: f(x)=x 3 –3x 2 +2x–5 or by putting x=y+1, y 3 –y–5 = 0, ie D= 4-2725<0, поэтому многочлен f(x) has one real root.
2. For a polynomial g(x) we determine the number of real roots by setting the number of sign changes in the Sturm system of the polynomial g(x) when going from –∞ to + ∞. We will also find whole boundaries between which each of these roots is located, and we will not build a graph of this function in advance.
Any polynomial g(x) with real coefficients and having no multiple roots possesses the Sturm system. If the polynomial has multiple roots, then you need to get rid of them by dividing the polynomial g(x) on the GCD of the polynomials g(x) and g"(x). The Sturm system of the polynomial g(x) can be constructed as follows: put g 1 (x)=g"(x), then divide g(x) on the g 1 (x) and the remainder of this division, taken with the opposite sign, is taken as g 2 (x), i.e. g(x)=g 1 (x)h 1 (x)–g 2 (x). In general, if the polynomials g k – 1 ( x) and g To ( x) have already been found, then g k + 1 ( x) will be the remainder of the division g k – 1 ( x) on the g To ( x), taken with the opposite sign:
g k – 1 ( x)=g To ( x)q To ( x)– g k + 1 ( x).
Find the Sturm system for g(x) using the specified method. In this case, in the process of division, we will, in contrast to the Euclidean algorithm, multiply and reduce only by arbitrary positive numbers, since residue marks play an important role in Sturm's method. We will get such a system
g(x)=x 3 +3x 2 –1,
g 1 (x)=3x 2 +6x,
g 2 (x)=2x+1,
g 3 (x)=1.
Let us determine the signs of the polynomials of this system for x= –∞ and x= + ∞, for which we look only at the signs of the leading coefficients and at the degrees of these polynomials. At + ∞, the signs of all polynomials of the Sturm system will coincide with the signs of their leading terms, and at –∞ the signs of the polynomials of the Sturm system coincide with the signs of their leading coefficients for polynomials of even degree and are opposite to the signs of the highest polynomials of odd degree.
Thus, when passing x from –∞ to + ∞, the Sturm system loses three sign changes, so the polynomial g(x) has exactly three real roots (Sturm's theorem).
Let us continue the study of signs in the Sturm system, considering the intervals (0,1), (1,2), (2,3), etc., (0, –1), (–1, –2), (–2 , –3), etc. Thus, we define the intervals ( but, b), where a – b= 1 containing three real roots and find the interval for x 0 .
Thus, the Sturm system of the polynomial g(x) loses one change of signs at the transition x–3 to –2, –1 to 0, and 0 to 1. Roots x 1 , x 2 , x 3 of this polynomial satisfy, therefore, the inequalities:
–3<x 1 <–2, –1<x 2 <0, 0<x 3 <1, т.е. наибольший корень x 0 (0,1).
3. Let us construct in the interval (0, 1) schematically the graph of the polynomial g(x) by calculating the following values of the polynomials:
g(0)=–1, g(1)=3, g"(0)=0, g"(1) = 9 (the function increases on the considered interval), g""(0)>0g"" (1)> 0 (convex function).
A schematic graph of the function is shown in Fig. 1.
First, by the method of chords on the segment (0,1), the curve y=g(x) is replaced by the chord AB and the abscissa is taken as the first approximate value of the root x= from the point of intersection of this chord with the axis x... The PBC triangle is similar to the CAE triangle, therefore, or, or. In general .
Then, using Newton's method, we draw the tangent y to the schedule g(x) at point A (1, g(1)) (we draw a tangent at the point x= 1, because g(1) and g"" (1) of the same sign) and take the abscissa for another approximate value of the root x=R the point of intersection of this tangent with the Ox axis.
We write the equation of the tangent line passing through point A
y–g(1)=g"(1)(x–1).
Since this tangent passes through the point ( p, 0), then substituting these values into the tangent equation, we obtain
0–g(1)=g"(1)(p–1) or p=1– =1– .
In general p=b– .
More precise value of the required root x 0 now you can search in the new
interval ( but 1 , b 1) by setting but 1 =0,3, b 1 = 0.7. Repeating the chord method and Newton's method in the interval ( but 1 , b 1) we have: g(but 1)=–0,703; g(b 1)=0,813; g "(b 1)=5,67.
As g(but 1) and g(b 1) different signs, then x 0 (but 1 ,b 1)
p 1 =0,7– .
Consider a new interval ( but 2 , b 2) by setting but 2 =0,5, b 2 =0,55, g(but 2)=–0,125, g(b 2)=0,073875, g "(b 2) = 4.2075, because g(but 2) and g(b 2) - different signs, then x 0 (but 2 ,b 2),
, p 2 =0,55– .
And finally, considering the interval ( but 3 , b 3), where but 3 =0,531, b 3 = 0.532, we find more precisely x 0 .
A task 18. The next rational fraction, where
f(x)= 2x 4 –10x 3 +7x 2 +4x+3, g(x)= x 5 –2x 3 +2x 2 –3x+2,
decompose into the sum of the simplest fractions in the field of rational numbers.
Solution. Every regular rational fraction has a unique expansion in the sum of the simplest fractions. In our case, the degree f(x) less degree g(x), so the fraction is correct.
