8 examples of factorization of polynomials are given. They include examples with solving squares and biquadratic equations, examples with recurrent polynomials, and examples with finding integer roots of third and fourth degree polynomials.
1. Examples with the solution of a quadratic equation
Example 1.1
x 4 + x 3 - 6 x 2.
Solution
Take out x 2
for brackets:
.
2 + x - 6 = 0:
.
Equation roots:
, .
.
Answer
Example 1.2
Factoring a third-degree polynomial:
x 3 + 6 x 2 + 9 x.
Solution
We take x out of brackets:
.
We decide quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant is .
Since the discriminant zero, then the roots of the equation are multiple: ;
.
From here we obtain the decomposition of the polynomial into factors:
.
Answer
Example 1.3
Factoring a fifth-degree polynomial:
x 5 - 2 x 4 + 10 x 3.
Solution
Take out x 3
for brackets:
.
We solve the quadratic equation x 2 - 2 x + 10 = 0.
Its discriminant is .
Since the discriminant is less than zero, the roots of the equation are complex: ;
, .
The factorization of a polynomial has the form:
.
If we are interested in factoring with real coefficients, then:
.
Answer
Examples of factoring polynomials using formulas
Examples with biquadratic polynomials
Example 2.1
Factorize the biquadratic polynomial:
x 4 + x 2 - 20.
Solution
Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).
;
.
Answer
Example 2.2
Factoring a polynomial that reduces to a biquadratic:
x 8 + x 4 + 1.
Solution
Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):
;
;
.
Answer
Example 2.3 with recursive polynomial
Factoring the recursive polynomial:
.
Solution
The recursive polynomial has an odd degree. Therefore it has a root x = - 1
. We divide the polynomial by x - (-1) = x + 1. As a result, we get:
.
We make a substitution:
, ;
;
;
.
Answer
Examples of Factoring Polynomials with Integer Roots
Example 3.1
Factoring a polynomial:
.
Solution
Suppose the equation
6
-6, -3, -2, -1, 1, 2, 3, 6
.
(-6) 3 - 6 (-6) 2 + 11 (-6) - 6 = -504;
(-3) 3 - 6 (-3) 2 + 11 (-3) - 6 = -120;
(-2) 3 - 6 (-2) 2 + 11 (-2) - 6 = -60;
(-1) 3 - 6 (-1) 2 + 11 (-1) - 6 = -24;
1 3 - 6 1 2 + 11 1 - 6 = 0;
2 3 - 6 2 2 + 11 2 - 6 = 0;
3 3 - 6 3 2 + 11 3 - 6 = 0;
6 3 - 6 6 2 + 11 6 - 6 = 60.
So, we have found three roots:
x 1 = 1
, x 2 = 2
, x 3 = 3
.
Since the original polynomial is of the third degree, it has no more than three roots. Since we have found three roots, they are simple. Then
.
Answer
Example 3.2
Factoring a polynomial:
.
Solution
Suppose the equation
has at least one integer root. Then it is the divisor of the number 2
(a member without x ). That is, the whole root can be one of the numbers:
-2, -1, 1, 2
.
Substitute these values one by one:
(-2) 4 + 2 (-2) 3 + 3 (-2) 3 + 4 (-2) + 2 = 6
;
(-1) 4 + 2 (-1) 3 + 3 (-1) 3 + 4 (-1) + 2 = 0
;
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 = 12;
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 = 54
.
If we assume that this equation has an integer root, then it is a divisor of the number 2
(a member without x ). That is, the whole root can be one of the numbers:
1, 2, -1, -2
.
Substitute x = -1
:
.
So we have found another root x 2
= -1
. It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.
Since the equation x 2 + 2 = 0 has no real roots, then the factorization of the polynomial has the form.
Factoring a polynomial. Part 1
Factorization is a universal technique that helps to solve complex equations and inequalities. The first thought that should come to mind when solving equations and inequalities in which the right side is zero is to try to factorize the left side.
We list the main ways to factorize a polynomial:
- taking the common factor out of the bracket
- use of abbreviated multiplication formulas
- by the formula for factoring a square trinomial
- grouping method
- dividing a polynomial by a binomial
- method of indeterminate coefficients
In this article we will dwell on the first three methods in detail, the rest will be discussed in the following articles.
1. Taking the common factor out of the bracket.
To take the common factor out of the bracket, you must first find it. Common multiplier coefficient is equal to the greatest common divisor of all coefficients.
Letter part the common factor is equal to the product of the expressions that make up each term with the smallest exponent.
The scheme for taking out a common factor looks like this:
Attention!
The number of terms in brackets is equal to the number of terms in the original expression. If one of the terms coincides with the common factor, then when it is divided by the common factor, we get one.
Example 1
Factorize the polynomial:
Let's take the common factor out of brackets. To do this, we first find it.
1. Find the largest common divisor all coefficients of the polynomial, i.e. numbers 20, 35 and 15. It is equal to 5.
2. We establish that the variable is contained in all terms, and the smallest of its exponents is 2. The variable is contained in all terms, and the smallest of its exponents is 3.
The variable is contained only in the second term, so it is not part of the common factor.
So the common factor is
3. We take out the factor using the scheme above:
Example 2 Solve the equation:
Solution. Let's factorize the left side of the equation. Let's take the factor out of brackets:
So we got the equation 
Set each factor equal to zero:
We get - the root of the first equation.
Roots:
Answer: -1, 2, 4
2. Factorization using abbreviated multiplication formulas.
If the number of terms in the polynomial that we are going to factorize is less than or equal to three, then we try to apply the reduced multiplication formulas.
1. If the polynomial isdifference of two terms, then we try to apply difference of squares formula:
or cube difference formula:
Here are the letters and denote a number or an algebraic expression.
2. If the polynomial is the sum of two terms, then perhaps it can be factored using formulas for the sum of cubes:
3. If the polynomial consists of three terms, then we try to apply sum square formula:
or difference square formula:
Or we try to factorize by formula for factoring a square trinomial:
Here and are the roots of the quadratic equation 
Example 3Factoring the expression:
Solution. We have the sum of two terms. Let's try to apply the formula for the sum of cubes. To do this, you must first represent each term as a cube of some expression, and then apply the formula for the sum of cubes:
Example 4 Factoring the expression: 
Solution. Before us is the difference of the squares of two expressions. First expression: , second expression:
Let's apply the formula for the difference of squares:
Let's open the brackets and give like terms, we get:
We already know how to partially use the factorization of the difference of degrees - when studying the topic “Difference of Squares” and “Difference of Cubes”, we learned to represent as a product the difference of expressions that can be represented as squares or as cubes of some expressions or numbers.
Abbreviated multiplication formulas
According to the formulas of abbreviated multiplication:
the difference of squares can be represented as the product of the difference of two numbers or expressions by their sum
The difference of cubes can be represented as the product of the difference of two numbers by the incomplete square of the sum
Transition to the difference of expressions in 4 powers
Based on the difference of squares formula, let's try to factorize the expression $a^4-b^4$
Recall how a power is raised to a power - for this, the base remains the same, and the exponents are multiplied, i.e. $((a^n))^m=a^(n*m)$
Then you can imagine:
$a^4=(((a)^2))^2$
$b^4=(((b)^2))^2$
So our expression can be represented as $a^4-b^4=(((a)^2))^2$-$(((b)^2))^2$
Now in the first bracket we again got the difference of numbers, which means we can again factorize as the product of the difference of two numbers or expressions by their sum: $a^2-b^2=\left(a-b\right)(a+b)$.
Now we calculate the product of the second and third brackets using the rule for the product of polynomials - we multiply each term of the first polynomial by each term of the second polynomial and add the result. To do this, we first multiply the first term of the first polynomial - $a$ - by the first and second terms of the second one (by $a^2$ and $b^2$), i.e. we get $a\cdot a^2+a\cdot b^2$, then we multiply the second term of the first polynomial -$b$- by the first and second terms of the second polynomial (by $a^2$ and $b^2$), those. get $b\cdot a^2 + b\cdot b^2$ and sum the resulting expressions
$\left(a+b\right)\left(a^2+b^2\right)=a\cdot a^2+a\cdot b^2+ b \cdot a^2 + b\cdot b^ 2 = a^3+ab^2+a^2b+b^3$
We write the difference of monomials of the 4th degree, taking into account the calculated product:
$a^4-b^4=(((a)^2))^2$-$(((b)^2))^2=((a)^2-b^2)(a^2 +b^2)$=$\ \left(ab\right)(a+b)(a^2+b^2)\ $=
Transition to the difference of expressions in the 6th power
Based on the difference of squares formula, let's try to factorize the expression $a^6-b^6$
Recall how a power is raised to a power - for this, the base remains the same, and the exponents are multiplied, i.e. $((a^n))^m=a^(n\cdot m)$
Then you can imagine:
$a^6=(((a)^3))^2$
$b^6=(((b)^3))^2$
So our expression can be represented as $a^6-b^6=(((a)^3))^2-(((b)^3))^2$
In the first bracket we got the difference of cubes of monomials, in the second the sum of cubes of monomials, now we can again factorize the difference of cubes of monomials as the product of the difference of two numbers by the incomplete square of the sum $a^3-b^3=\left(ab\right)( a^2+ab+b^2)$
The original expression takes the form
$a^6-b^6=((a)^3-b^3)\left(a^3+b^3\right)=\left(ab\right)(a^2+ab+b^ 2)(a^3+b^3)$
We calculate the product of the second and third brackets using the rule for the product of polynomials - we multiply each term of the first polynomial by each term of the second polynomial and add the result.
$(a^2+ab+b^2)(a^3+b^3)=a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$
We write the difference of monomials of the 6th degree, taking into account the calculated product:
$a^6-b^6=((a)^3-b^3)\left(a^3+b^3\right)=\left(ab\right)(a^2+ab+b^ 2)(a^3+b^3)=(ab)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)$
Factoring the power difference
Let us analyze the formulas for the difference of cubes, the difference of $4$ degrees, the difference of $6$ degrees
We see that in each of these expansions there is some analogy, generalizing which we get:
Example 1
Factorize $(32x)^(10)-(243y)^(15)$
Solution: First, we represent each monomial as some monomial to the power of 5:
\[(32x)^(10)=((2x^2))^5\]\[(243y)^(15)=((3y^3))^5\]
We use the power difference formula
Picture 1.
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Very often the numerator and denominator of a fraction are algebraic expressions, which first need to be decomposed into factors, and then, having found the same among them, divide both the numerator and the denominator into them, that is, reduce the fraction. A whole chapter of a textbook on algebra in the 7th grade is devoted to tasks to factorize a polynomial. Factoring can be done 3 ways, as well as a combination of these methods.
1. Application of abbreviated multiplication formulas
As is known to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and add the resulting products. There are at least 7 (seven) common cases of multiplication of polynomials that are included in the concept. For example,
Table 1. Factorization in the 1st way
2. Taking the common factor out of the bracket
This method is based on the application of the distributive law of multiplication. For example,
We divide each term of the original expression by the factor that we take out, and at the same time we get the expression in brackets (that is, the result of dividing what was by what we take out remains in brackets). First of all, you need correctly determine the multiplier, which must be bracketed.
The polynomial in brackets can also be a common factor:
When performing the “factorize” task, one must be especially careful with the signs when taking the common factor out of brackets. To change the sign of each term in a parenthesis (b - a), we take out the common factor -1 , while each term in the bracket is divided by -1: (b - a) = - (a - b) .
In the event that the expression in brackets is squared (or to any even power), then numbers inside brackets can be swapped completely free, since the minuses taken out of brackets will still turn into a plus when multiplied: (b - a) 2 = (a - b) 2, (b - a) 4 = (a - b) 4 etc…
3. Grouping method
Sometimes not all terms in the expression have a common factor, but only some. Then you can try group terms in brackets so that some factor can be taken out of each. Grouping method is double bracketing of common factors.
4. Using several methods at once
Sometimes you need to apply not one, but several ways to factorize a polynomial into factors at once.
This is a synopsis on the topic. "Factorization". Choose next steps:
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