Factoring polynomials is an identity transformation, as a result of which a polynomial is transformed into the product of several factors - polynomials or monomials.
There are several ways to factor polynomials.
Method 1. Taking the common factor out of brackets.
This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to isolate the common factor in the two components under consideration and “take” it out of brackets.
Let us factor the polynomial 28x 3 – 35x 4.
Solution.
1. Find the elements 28x 3 and 35x 4 common divisor. For 28 and 35 it will be 7; for x 3 and x 4 – x 3. In other words, our common factor is 7x 3.
2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x.
3. We take the common factor out of brackets
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x = 7x 3 (4 – 5x).
Method 2. Using abbreviated multiplication formulas. The “mastery” of using this method is to notice one of the abbreviated multiplication formulas in the expression.
Let us factor the polynomial x 6 – 1.
Solution.
1. We can apply the difference of squares formula to this expression. To do this, imagine x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1).
2. We can apply the formula for the sum and difference of cubes to the resulting expression:
(x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).
So,
x 6 – 1 = (x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).
Method 3. Grouping. The grouping method is to combine the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, subtraction of a common factor).
Let's factor the polynomial x 3 – 3x 2 + 5x – 15.
Solution.
1. Let's group the components in this way: the 1st with the 2nd, and the 3rd with the 4th
(x 3 – 3x 2) + (5x – 15).
2. In the resulting expression, we take the common factors out of brackets: x 2 in the first case and 5 in the second.
(x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3).
3. We take the common factor x – 3 out of brackets and get:
x 2 (x – 3) + 5(x – 3) = (x – 3)(x 2 + 5).
So,
x 3 – 3x 2 + 5x – 15 = (x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3) = (x – 3) ∙ (x 2 + 5 ).
Let's secure the material.
Factor the polynomial a 2 – 7ab + 12b 2 .
Solution.
1. Let us represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 – (3ab + 4ab) + 12b 2. 
Let's open the brackets and get:
a 2 – 3ab – 4ab + 12b 2.
2. Let's group the components of the polynomial in this way: 1st with 2nd and 3rd with 4th. We get:
(a 2 – 3ab) – (4ab – 12b 2).
3. Let’s take the common factors out of brackets:
(a 2 – 3ab) – (4ab – 12b 2) = a(a – 3b) – 4b(a – 3b).
4. Let’s take the common factor (a – 3b) out of brackets:
a(a – 3b) – 4b(a – 3b) = (a – 3 b) ∙ (a – 4b).
So,
a 2 – 7ab + 12b 2 =
= a 2 – (3ab + 4ab) + 12b 2 =
= a 2 – 3ab – 4ab + 12b 2 =
= (a 2 – 3ab) – (4ab – 12b 2) =
= a(a – 3b) – 4b(a – 3b) =
= (a – 3 b) ∙ (a – 4b).
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Factoring polynomials is an identity transformation, as a result of which a polynomial is transformed into the product of several factors - polynomials or monomials.
There are several ways to factor polynomials.
Method 1. Taking the common factor out of brackets.
This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to isolate the common factor in the two components under consideration and “take” it out of brackets.
Let us factor the polynomial 28x 3 – 35x 4.
Solution.
1. Find a common divisor for the elements 28x3 and 35x4. For 28 and 35 it will be 7; for x 3 and x 4 – x 3. In other words, our common factor is 7x 3.
2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x.
3. We take the common factor out of brackets
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x = 7x 3 (4 – 5x).
Method 2. Using abbreviated multiplication formulas. The “mastery” of using this method is to notice one of the abbreviated multiplication formulas in the expression.
Let us factor the polynomial x 6 – 1.
Solution.
1. We can apply the difference of squares formula to this expression. To do this, imagine x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1).
2. We can apply the formula for the sum and difference of cubes to the resulting expression:
(x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).
So,
x 6 – 1 = (x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).
Method 3. Grouping. The grouping method is to combine the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, subtraction of a common factor).
Let's factor the polynomial x 3 – 3x 2 + 5x – 15.
Solution.
1. Let's group the components in this way: the 1st with the 2nd, and the 3rd with the 4th
(x 3 – 3x 2) + (5x – 15).
2. In the resulting expression, we take the common factors out of brackets: x 2 in the first case and 5 in the second.
(x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3).
3. We take the common factor x – 3 out of brackets and get:
x 2 (x – 3) + 5(x – 3) = (x – 3)(x 2 + 5).
So,
x 3 – 3x 2 + 5x – 15 = (x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3) = (x – 3) ∙ (x 2 + 5 ).
Let's secure the material.
Factor the polynomial a 2 – 7ab + 12b 2 .
Solution.
1. Let us represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 – (3ab + 4ab) + 12b 2.
Let's open the brackets and get:
a 2 – 3ab – 4ab + 12b 2.
2. Let's group the components of the polynomial in this way: 1st with 2nd and 3rd with 4th. We get:
(a 2 – 3ab) – (4ab – 12b 2).
3. Let’s take the common factors out of brackets:
(a 2 – 3ab) – (4ab – 12b 2) = a(a – 3b) – 4b(a – 3b).
4. Let’s take the common factor (a – 3b) out of brackets:
a(a – 3b) – 4b(a – 3b) = (a – 3 b) ∙ (a – 4b).
So,
a 2 – 7ab + 12b 2 =
= a 2 – (3ab + 4ab) + 12b 2 =
= a 2 – 3ab – 4ab + 12b 2 =
= (a 2 – 3ab) – (4ab – 12b 2) =
= a(a – 3b) – 4b(a – 3b) =
= (a – 3 b) ∙ (a – 4b).
website, when copying material in full or in part, a link to the source is required.
A polynomial is an expression consisting of the sum of monomials. The latter are the product of a constant (number) and the root (or roots) of the expression to the power of k. In this case, we speak of a polynomial of degree k. The expansion of a polynomial involves a transformation of the expression in which the terms are replaced by factors. Let's consider the main ways to carry out this kind of transformation.
Method of expanding a polynomial by isolating a common factor
This method is based on the laws of the distribution law. So, mn + mk = m * (n + k).
- Example: expand 7y 2 + 2uy and 2m 3 – 12m 2 + 4lm.
7y 2 + 2uy = y * (7y + 2u),
2m 3 – 12m 2 + 4lm = 2m(m 2 – 6m + 2l).
However, the factor that is necessarily present in each polynomial may not always be found, therefore this method is not universal.
Polynomial expansion method based on abbreviated multiplication formulas
Abbreviated multiplication formulas are valid for polynomials of any degree. IN general view The conversion expression looks like this:
u k – l k = (u – l)(u k-1 + u k-2 * l + u k-3 *l 2 + … u * l k-2 + l k-1), where k is a representative of natural numbers .
The formulas most often used in practice are for polynomials of the second and third orders:
u 2 – l 2 = (u – l)(u + l),
u 3 – l 3 = (u – l)(u 2 + ul + l 2),
u 3 + l 3 = (u + l)(u 2 – ul + l 2).
- Example: expand 25p 2 – 144b 2 and 64m 3 – 8l 3.
25p 2 – 144b 2 = (5p – 12b)(5p + 12b),
64m 3 – 8l 3 = (4m) 3 – (2l) 3 = (4m – 2l)((4m) 2 + 4m * 2l + (2l) 2) = (4m – 2l)(16m 2 + 8ml + 4l 2 ).
Polynomial expansion method - grouping terms of an expression
This method in some way has something in common with the technique of deriving the common factor, but has some differences. In particular, before isolating a common factor, the monomials should be grouped. The grouping is based on the rules of combinational and commutative laws.
All monomials presented in the expression are divided into groups, in each of which general meaning such that the second factor will be the same in all groups. In general, this decomposition method can be represented as the expression:
pl + ks + kl + ps = (pl + ps) + (ks + kl) ⇒ pl + ks + kl + ps = p(l + s) + k(l + s),
pl + ks + kl + ps = (p + k)(l + s).
- Example: spread out 14mn + 16ln – 49m – 56l.
14mn + 16ln – 49m – 56l = (14mn – 49m) + (16ln – 56l) = 7m * (2n – 7) + 8l * (2n – 7) = (7m + 8l)(2n – 7).
Polynomial expansion method - forming a perfect square
This method is one of the most effective in the expansion of a polynomial. At the initial stage, it is necessary to determine monomials that can be “collapsed” into the square of the difference or sum. To do this, use one of the relations:
(p – b) 2 = p 2 – 2pb + b 2,
- Example: expand the expression u 4 + 4u 2 – 1.
Among its monomials, we select the terms that form a complete square: u 4 + 4u 2 – 1 = u 4 + 2 * 2u 2 + 4 – 4 – 1 =
= (u 4 + 2 * 2u 2 + 4) – 4 – 1 = (u 4 + 2 * 2u 2 + 4) – 5.
Complete the transformation using the abbreviated multiplication rules: (u 2 + 2) 2 – 5 = (u 2 + 2 – √5)(u 2 + 2 + √5).
That. u 4 + 4u 2 – 1 = (u 2 + 2 – √5)(u 2 + 2 + √5).
In order to factorize, it is necessary to simplify the expressions. This is necessary so that it can be further reduced. The expansion of a polynomial makes sense when its degree is not lower than two. A polynomial with the first degree is called linear.
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The article will cover all the concepts of decomposition, theoretical basis and methods of factoring a polynomial.
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 degree a n and n linear factors (x - x i), i = 1, 2, ..., n, then P n (x) = a n (x - x n) (x - x n - 1) · . . . · (x - x 1) , where x i, i = 1, 2, …, n 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 the complex roots that will occur in conjugate pairs. For example, roots x 1 and x 2 related 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 complex conjugate, then the other roots are real, from which 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 a polynomial can be repeated. Let's consider the proof of the algebra theorem, a consequence of Bezout's theorem.
Fundamental theorem of algebra
Theorem 2Any polynomial with degree n has at least one root.
Bezout's theorem
After dividing a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 on (x - s), then we get the remainder, which is equal to the polynomial at 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 with degree n - 1.
Corollary to Bezout's theorem
When the root of the polynomial P n (x) is considered to be 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 the solution.
Factoring a quadratic trinomial
A square trinomial of the form a x 2 + b x + c can be factorized 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 is clear that the expansion itself reduces to the solution quadratic equation subsequently.
Example 1
Factor the quadratic trinomial.
Solution
It is necessary to 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 using the formula, then we get D = (- 5) 2 - 4 · 4 · 1 = 9. From here 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.
To perform the check, you need to open the parentheses. 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 arrive at the original expression. That is, we can conclude that the decomposition was performed correctly.
Example 2
Factor the quadratic trinomial of the form 3 x 2 - 7 x - 11 .
Solution
We find 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 we 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 the expansion itself can be depicted as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.
Example 4
Decompose the quadratic 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 i x 2 = - 1 3 - D 2 1 = - 1 3 - 35 3 i 2 = - 1 - 35 i 6 = - 1 6 - 35 6 i
Having obtained 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 discriminant value is negative, then the polynomials will remain second-order polynomials. It follows from this that we will not expand them into linear factors.
Methods for factoring a polynomial of degree higher than two
When decomposing, a universal method is assumed. 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 reduce 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 obtain a complete expansion.
If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic involves solving equations with higher degrees and integer coefficients.
Taking the common factor out of brackets
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 to be taking the common factor out of brackets.
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 remove x from the brackets of the entire expression. We get:
4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)
Let's move on 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 x x - - 1 + 5 2 x - - 1 - 5 2 = = 4 x x + 1 - 5 2 x + 1 + 5 2
To begin with, let us take into consideration a 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 of the highest degree is 1.
When a polynomial has integer roots, then they are considered divisors of the free term.
Example 6
Decompose the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.
Solution
Let's consider whether there are complete 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 integer roots. You can check using Horner's scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a 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 proceed to the expansion of a quadratic trinomial of the form x 2 + 2 x + 3.
Since the discriminant is negative, it means 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 root selection and division of a polynomial by a polynomial instead of Horner's scheme. Let's move on to considering 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 highest of which is equal to one.
This case occurs for rational fractions.
Example 7
Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .
Solution
It is necessary to replace the variable y = 2 x, you should move on 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 their location is among the divisors of the free term. The entry will look like:
±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60
Let's move on to calculating the function g (y) at these points in order to get 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 find that y = - 5 is the root of an equation of the form y 3 + 19 y 2 + 82 y + 60, which means 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 it down 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 factorize the resulting quadratic trinomial of the form x 2 + 7 x + 3. By equating to zero we 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
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 techniques for factoring a polynomial
Rational roots are not inherent in all polynomials. To do this, you need to use special methods to find factors. But not all polynomials can be expanded or represented as a product.
Grouping method
There are cases when you can group the terms of a polynomial to find a common factor and put it out of brackets.
Example 9
Factor the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.
Solution
Because the coefficients are integers, then the roots can presumably 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
This shows that there are no roots; it is necessary to use another method of expansion 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 factorize. 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 grouping does not mean that choosing terms is easy enough. There is no specific solution method, 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 integer roots. The terms should be grouped. 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 factorization 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 to factor a polynomial
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 Newton’s 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 sequence of coefficients of the sum in parentheses is indicated by the expression x + 1 4 .
This means 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 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 that is in the second bracket. It is clear that there are no knights there, so we should apply the difference of squares formula 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
Factorize x 3 + 6 x 2 + 12 x + 6 .
Solution
Let's start transforming 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 method for replacing a variable when factoring a polynomial
When replacing a variable, the degree is reduced and the polynomial is factored.
Example 13
Factor the 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 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 obtained the desired decomposition.
The cases discussed above will help in considering and factoring a polynomial in different ways.
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