In this lesson, we will recall all the previously studied methods of factoring a polynomial and consider examples of their application, in addition, we will study a new method - the full square method and learn how to apply it in solving various problems.
Subject:Factoring polynomials
Lesson:Factorization of polynomials. Full square selection method. Combination of methods
Recall the main methods for factoring a polynomial that were studied earlier:
The method of taking a common factor out of brackets, that is, a factor that is present in all members of the polynomial. Consider an example:
Recall that a monomial is a product of powers and numbers. In our example, both members have some common, identical elements.
So, let's take the common factor out of brackets:
;
Recall that by multiplying the rendered multiplier by the bracket, you can check the correctness of the rendering.
grouping method. It is not always possible to take out a common factor in a polynomial. In this case, you need to divide its members into groups in such a way that in each group you can take out a common factor and try to break it down so that after taking out the factors in the groups, a common factor appears for the entire expression, and expansion could be continued. Consider an example:
Group the first term with the fourth, the second with the fifth, and the third with the sixth respectively:
Let's take out the common factors in the groups:
The expression has a common factor. Let's take it out:
Application of abbreviated multiplication formulas. Consider an example:
;
Let's write the expression in detail:
Obviously, we have before us the formula for the square of the difference, since there is a sum of the squares of two expressions and their double product is subtracted from it. Let's roll by the formula:
Today we will learn another way - the full square selection method. It is based on the formulas of the square of the sum and the square of the difference. Recall them:
The formula for the square of the sum (difference);
The peculiarity of these formulas is that they contain squares of two expressions and their double product. Consider an example:
Let's write the expression:
So the first expression is , and the second .
In order to make a formula for the square of the sum or difference, the double product of the expressions is not enough. It needs to be added and subtracted:
Let's collapse the full square of the sum:
Let's transform the resulting expression:
We apply the difference of squares formula, recall that the difference of the squares of two expressions is the product and the sums by their difference:
So, this method consists, first of all, in the fact that it is necessary to identify the expressions a and b that are squared, that is, to determine which expressions are squared in this example. After that, you need to check for the presence of a double product and if it is not there, then add and subtract it, this will not change the meaning of the example, but the polynomial can be factored using the formulas for the square of the sum or difference and difference of squares, if possible.
Let's move on to solving examples.
Example 1 - factorize:
Find expressions that are squared:
Let's write down what their double product should be:
Let's add and subtract the double product:
Let's collapse the full square of the sum and give similar ones:
We will write according to the formula of the difference of squares:
Example 2 - solve the equation:
;
There is a trinomial on the left side of the equation. You need to factor it out. We use the formula of the square of the difference:
We have the square of the first expression and the double product, the square of the second expression is missing, let's add and subtract it:
Let us collapse the full square and give like terms:
Let's apply the difference of squares formula:
So we have the equation 
We know that the product is equal to zero only if at least one of the factors is equal to zero. Based on this, we will write equations:
Let's solve the first equation:
Let's solve the second equation:
Answer: or
;
We act similarly to the previous example - select the square of the difference.
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 not lower than the second. A polynomial with the first degree is called linear.
Yandex.RTB R-A-339285-1
The article will reveal all the concepts of decomposition, theoretical foundations and methods for 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 - 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 will occur in conjugate pairs. For example, the 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, hence we get 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. Consider the proof of the theorem of algebra, the consequences of Bezout's theorem.
Fundamental theorem of algebra
Theorem 2Any polynomial with degree n has at least one root.
Bezout's theorem
After dividing the 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 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 with degree n - 1 .
Corollary from 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.
Factorization of a square trinomial
A square trinomial of the form a x 2 + b x + c can be factored into linear factors. then we get that a x 2 + b x + c \u003d a (x - x 1) (x - x 2) , where x 1 and x 2 are roots (complex or real).
This shows that the decomposition itself reduces to solving the quadratic equation later.
Example 1
Factorize a square 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 according to the formula, then we get D \u003d (- 5) 2 - 4 4 1 \u003d 9. Hence we have that
x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1
From here we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.
To perform the check, you need to open the brackets. 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 verification, we arrive at the original expression. That is, we can conclude that the expansion is correct.
Example 2
Factorize 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 - 1816
From here we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6 .
Example 3
Factorize 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
Expand 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 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 value of the discriminant is negative, then the polynomials will remain second-order polynomials. Hence it follows that we will not decompose them into linear factors.
Methods for factoring a polynomial of degree higher than the second
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 powers 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 \u003d 0, then you can represent the polynomial in the form of an expression P n (x) \u003d 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
Factorize the third degree polynomial 4 x 3 + 8 x 2 - x.
Solution
We see that x 1 \u003d 0 is the root of the given polynomial, then we can bracket x out 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 the 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's take for 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 power is 1 .
When the polynomial has integer 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 whether there are integer roots. It is necessary to write out 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 according to the Horner scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a polynomial:
It follows that x \u003d 2 and x \u003d - 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 turn 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) \u003d x 4 + 3 x 3 - x 2 - 9 x - 18 \u003d (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 us 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 highest of which does not equal one.
This case takes place for fractional rational fractions.
Example 7
Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .
Solution
It is necessary to change the variable y = 2 x , one should pass 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) \u003d y 3 + 19 y 2 + 82 y + 60 has integer roots, then their finding 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 proceed to the calculation of 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 get that y \u003d - 5 is the root of the equation of the form y 3 + 19 y 2 + 82 y + 60, which means that x \u003d y 2 \u003d - 5 2 is the root of the original function.
Example 8
It is necessary to divide by a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.
Solution
We 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. 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
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 when 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 decomposed 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 take it out of 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 can presumably also be integers. To check, we 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 a different method of decomposition and solution.
Grouping is required:
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 a 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 it is easy enough to choose terms. There is no definite way to solve it, therefore it is necessary to use special theorems and rules.
Example 10
Factorize 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 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 and Newton's binomial formulas to factorize a polynomial
Appearance often does not always make it clear which way to use 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
Factorize 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 brackets is indicated by the expression x + 1 4 .
So 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 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 like
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 change 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 changing a variable, the degree is reduced and the polynomial is factorized.
Example 13
Factorize a polynomial of the form x 6 + 5 x 3 + 6 .
Solution
By the condition, it is clear that it is necessary to make a 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 the 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 have obtained the desired expansion.
The cases discussed above will help in considering and factoring a polynomial in various ways.
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The factorization of polynomials is an identical transformation, as a result of which a polynomial is transformed into a product of several factors - polynomials or monomials.
There are several ways to factorize polynomials.
Method 1. Bracketing the common factor.
This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to single out the common factor in the two components under consideration and “put it out” of the brackets.
Let us factorize the polynomial 28x 3 - 35x 4.
Solution.
1. We find a common divisor for 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 7x3.
2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x.
3. Bracketing the common factor
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x \u003d 7x 3 (4 - 5x).
Method 2. Using abbreviated multiplication formulas. The "mastery" of mastering this method is to notice in the expression one of the formulas for abbreviated multiplication.
Let us factorize the polynomial x 6 - 1.
Solution.
1. We can apply the difference of squares formula to this expression. To do this, we represent x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 - 1 \u003d (x 3 + 1) ∙ (x 3 - 1).
2. To the resulting expression, we can apply the formula for the sum and difference of cubes:
(x 3 + 1) ∙ (x 3 - 1) \u003d (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 consists in combining the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, taking out a common factor).
We factorize the polynomial x 3 - 3x 2 + 5x - 15.
Solution.
1. 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) \u003d x 2 (x - 3) + 5 (x - 3).
3. We take out the common factor x - 3 and get:
x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) (x 2 + 5).
So,
x 3 - 3x 2 + 5x - 15 \u003d (x 3 - 3x 2) + (5x - 15) \u003d x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) ∙ (x 2 + 5 ).
Let's fix the material.
Factor the polynomial a 2 - 7ab + 12b 2 .
Solution.
1. We 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. Group the components of the polynomial in this way: the 1st with the 2nd and the 3rd with the 4th. We get:
(a 2 - 3ab) - (4ab - 12b 2).
3. Let's take out the common factors:
(a 2 - 3ab) - (4ab - 12b 2) \u003d a (a - 3b) - 4b (a - 3b).
4. Let's take out the common factor (a - 3b):
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) =
= (а – 3 b) ∙ (а – 4b).
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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 k. In this case, one speaks of a polynomial of degree k. The decomposition of a polynomial involves the transformation of the expression, in which the terms are replaced by factors. Let us consider the main ways of carrying out this kind of transformation.
Method for expanding a polynomial by extracting 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, so this method is not universal.
Polynomial expansion method based on abbreviated multiplication formulas
Abbreviated multiplication formulas are valid for a polynomial of any degree. In general, the transformation 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 .
Most often in practice, formulas for polynomials of the second and third orders are used:
u 2 - l 2 \u003d (u - l) (u + l),
u 3 - l 3 \u003d (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 \u003d (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 decomposition method - grouping terms of an expression
This method in some way echoes the technique of deriving a common factor, but has some differences. In particular, before isolating the common factor, one should group the monomials. Grouping is based on the rules of associative and commutative laws.
All monomials presented in the expression are divided into groups, in each of which a common value is taken out such that the second factor will be the same in all groups. In general, such a decomposition method can be represented as an 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: expand 14mn + 16ln - 49m - 56l.
14mn + 16ln - 49m - 56l = (14mn - 49m) + (16ln - 56l) = 7m * (2n - 7) + 8l * (2n - 7) = (7m + 8l)(2n - 7).
Polynomial Decomposition Method - Full Square Formation
This method is one of the most efficient in the course of polynomial decomposition. At the initial stage, it is necessary to determine the monomials that can be “folded” into the square of the difference or sum. For this, one of the following relations is used:
(p - b) 2 \u003d p 2 - 2pb + b 2,
- Example: expand the expression u 4 + 4u 2 – 1.
Among its monomials, we single out the terms that form a complete square: u 4 + 4u 2 - 1 = u 4 + 2 * 2u 2 + 4 - 4 - 1 =
\u003d (u 4 + 2 * 2u 2 + 4) - 4 - 1 \u003d (u 4 + 2 * 2u 2 + 4) - 5.
Complete the transformation using the rules of abbreviated multiplication: (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).
