In the previous lesson we studied multiplying a polynomial by a monomial. For example, the product of a monomial a and a polynomial b + c is found as follows:

a(b + c) = ab + bc

However, in some cases it is more convenient to perform the inverse operation, which can be called taking the common factor out of brackets:

ab + bc = a(b + c)

For example, let us need to calculate the value of the polynomial ab + bc for the values ​​of the variables a = 15.6, b = 7.2, c = 2.8. If we substitute them directly into the expression, we get

ab + bc = 15.6 * 7.2 + 15.6 * 2.8

ab + bc = a(b + c) = 15.6 * (7.2 + 2.8) = 15.6 * 10 = 156

In this case, we represented the polynomial ab + bc as the product of two factors: a and b + c. This action is called factoring a polynomial.

Moreover, each of the factors into which the polynomial is expanded can, in turn, be a polynomial or a monomial.

Let's consider the polynomial 14ab - 63b 2. Each of its constituent monomials can be represented as a product:

It can be seen that both polynomials have a common factor 7b. This means that it can be taken out of brackets:

14ab - 63b 2 = 7b*2a - 7b*9b = 7b(2a-9b)

You can check whether the multiplier is correctly placed outside the brackets using the reverse operation - opening the brackets:

7b(2a - 9b) = 7b*2a - 7b*9b = 14ab - 63b 2

It is important to understand that often a polynomial can be expanded in several ways, for example:

5abc + 6bcd = b(5ac + 6cd) = c(5ab + 6bd) = bc(5a + 6d)

Usually they try to extract, roughly speaking, the “largest” monomial. That is, they expand the polynomial so that nothing more can be taken out of the remaining polynomial. So, during decomposition

5abc + 6bcd = b(5ac + 6cd)

the sum of monomials that have a common factor c remains in parentheses. If we take it out too, then there will be no common factors left in brackets:

b(5ac + 6cd) = bc(5a + 6d)

Let's look in more detail at how to find common factors of monomials. Let us decompose the sum

8a 3 b 4 + 12a 2 b 5 v + 16a 4 b 3 c 10

It consists of three terms. First, let's look at the numerical odds in front of them. These are 8, 12 and 16. In lesson 3 of 6th grade, the topic of GCD and the algorithm for finding it were discussed. This is the greatest common divisor. You can almost always find it orally. The numerical coefficient of the common multiplier will be exactly the GCD of the numerical coefficients of the terms of the polynomial. In this case, the number is 4.

Next, we look at the degrees of these variables. In a common factor, the letters must have the minimum powers that appear in the terms. So, the variable a in a polynomial has degrees 3, 2, and 4 (minimum 2), so the common factor will be a 2. The variable b has a minimum degree of 3, so the common factor will be b 3:

8a 3 b 4 + 12a 2 b 5 v + 16a 4 b 3 c 10 = 4a 2 b 3 (2ab + 3b 2 c + 4a 2 c 10)

As a result, the remaining terms 2ab, 3b 2 c, 4a 2 c 10 do not have a single common letter variable, and their coefficients 2, 3 and 4 do not have common divisors.

Not only monomials, but also polynomials can be taken out of brackets. For example:

x(a-5) + 2y(a-5) = (a-5)(x+2y)

Another example. It is necessary to expand the expression

5t(8y - 3x) + 2s(3x - 8y)

Solution. Recall that the minus sign reverses the signs in parentheses, so

-(8y - 3x) = -8y + 3x = 3x - 8y

This means we can replace (3x - 8y) with - (8y - 3x):

5t(8y - 3x) + 2s(3x - 8y) = 5t(8y - 3x) + 2*(-1)s(8y - 3x) = (8y - 3x)(5t - 2s)

Answer: (8y - 3x)(5t - 2s).

Remember that the subtrahend and minuend can be swapped by changing the sign in front of the brackets:

(a - b) = - (b - a)

The converse is also true: the minus sign already in front of the parentheses can be removed by simultaneously swapping the subtrahend and minuend:

This technique is often used when solving problems.

Grouping method

Let's consider another way to factor a polynomial, which helps to expand a polynomial. Let there be an expression

ab - 5a + bc - 5c

It is impossible to derive a factor common to all four monomials. However, you can imagine this polynomial as the sum of two polynomials, and in each of them take the variable out of brackets:

ab - 5a + bc - 5c = (ab - 5a) + (bc - 5c) = a(b - 5) + c(b - 5)

Now we can derive the expression b - 5:

a(b - 5) + c(b - 5) = (b - 5)(a + c)

We “grouped” the first term with the second, and the third with the fourth. Therefore, the described method is called the grouping method.

Example. Let us expand the polynomial 6xy + ab- 2bx- 3ay.

Solution. Grouping the 1st and 2nd terms is impossible, since they do not have a common factor. Therefore, let's swap the monomials:

6xy + ab - 2bx - 3ay = 6xy - 2bx + ab - 3ay = (6xy - 2bx) + (ab - 3ay) = 2x(3y - b) + a(b - 3y)

The differences 3y - b and b - 3y differ only in the order of the variables. In one of the brackets it can be changed by moving the minus sign out of the brackets:

(b - 3y) = - (3y - b)

Let's use this replacement:

2x(3y - b) + a(b - 3y) = 2x(3y - b) - a(3y - b) = (3y - b)(2x - a)

As a result, we got the identity:

6xy + ab - 2bx - 3ay = (3y - b)(2x - a)

Answer: (3y - b)(2x - a)

You can group not only two, but generally any number of terms. For example, in the polynomial

x 2 - 3xy + xz + 2x - 6y + 2z

we can group the first three and last 3 monomials:

x 2 - 3xy + xz + 2x - 6y + 2z = (x 2 - 3xy + xz) + (2x - 6y + 2z) = x(x - 3y + z) + 2(x - 3y + z) = (x + 2)(x - 3y + z)

Now let's look at a task of increased complexity

Example. Expand the quadratic trinomial x 2 - 8x +15.

Solution. This polynomial consists of only 3 monomials, and therefore, as it seems, grouping will not be possible. However, you can make the following replacement:

Then the original trinomial can be represented as follows:

x 2 - 8x + 15 = x 2 - 3x - 5x + 15

Let's group the terms:

x 2 - 3x - 5x + 15 = (x 2 - 3x) + (- 5x + 15) = x(x - 3) - 5(x - 3) = (x - 5)(x - 3)

Answer: (x- 5)(x - 3).

Of course, it is not easy to guess the replacement - 8x = - 3x - 5x in the above example. Let us show a different line of reasoning. We need to expand the polynomial of the second degree. As we remember, when multiplying polynomials, their powers add up. This means that even if we can factor a quadratic trinomial into two factors, they will turn out to be two polynomials of the 1st degree. Let us write the product of two polynomials of the first degree, whose leading coefficients are equal to 1:

(x + a)(x + b) = x 2 + xa + xb + ab = x 2 + (a + b)x + ab

Here we denote a and b as some arbitrary numbers. In order for this product to be equal to the original trinomial x 2 - 8x +15, it is necessary to select suitable coefficients for the variables:

Using selection, we can determine that the numbers a = - 3 and b = - 5 satisfy this condition. Then

(x - 3)(x - 5) = x 2 * 8x + 15

as can be seen by opening the brackets.

For simplicity, we considered only the case when the multiplied polynomials of the 1st degree have leading coefficients equal to 1. However, they could be equal, for example, to 0.5 and 2. In this case, the expansion would look slightly different:

x 2 * 8x + 15 = (2x - 6)(0.5x - 2.5)

However, taking coefficient 2 out of the first bracket and multiplying it by the second, we would get the original expansion:

(2x - 6)(0.5x - 2.5) = (x - 3) * 2 * (0.5x - 2.5) = (x - 3)(x - 5)

In the example considered, we expanded the quadratic trinomial into two polynomials of the first degree. We will have to do this often in the future. However, it is worth noting that some quadratic trinomials, e.g.

it is impossible to decompose in this way into a product of polynomials. This will be proven later.

Application of factoring polynomials

Factoring a polynomial can make some operations easier. Let it be necessary to calculate the value of the expression

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8 + 2 9

Let's take out the number 2, and the degree of each term will decrease by one:

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8 + 2 9 = 2(1 + 2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8)

Let's denote the amount

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8

for x. Then the equality written above can be rewritten:

x + 2 9 = 2(1 + x)

We got an equation, let’s solve it (see equation lesson):

x + 2 9 = 2(1 + x)

x + 2 9 = 2 + 2x

2x - x = 2 9 - 2

x = 512 - 2 = 510

Now let’s express the amount we are looking for in terms of x:

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8 + 2 9 = x + 2 9 = 510 + 512 = 1022

When solving this problem, we raised the number 2 only to the 9th power, and all other operations of exponentiation were eliminated from the calculations by factoring the polynomial. Similarly, you can create a calculation formula for other similar amounts.

Now let's calculate the value of the expression

38.4 2 - 61.6 * 29.5 + 61.6 * 38.4 - 29.5 * 38.4

38.4 2 - 61.6 * 29.5 + 61.6 * 38.4 - 29.5 * 38.4 = 38.4 2 - 29.5 * 38.4 + 61.6 * 38.4 - 61.6 * 29.5 = 38.4(38.4 - 29.5) + 61.6(38.4 - 29.5) = (38.4 + 61.6)(38.4 - 29.5) = 8.9*100 = 890

81 4 - 9 7 + 3 12

is divisible by 73. Note that the numbers 9 and 81 are powers of three:

81 = 9 2 = (3 2) 2 = 3 4

Knowing this, let's make a replacement in the original expression:

81 4 - 9 7 + 3 12 = (3 4) 4 - (3 2) 7 + 3 12 = 3 16 - 3 14 + 3 12

Let's take out 3 12:

3 16 - 3 14 + 3 12 = 3 12 (3 4 - 3 2 + 1) = 3 12 * (81 - 9 + 1) = 3 12 * 73

The product 3 12 .73 is divisible by 73 (since one of the factors is divisible by it), therefore the expression 81 4 - 9 7 + 3 12 is divided by this number.

Factoring can be used to prove identities. For example, let us prove the equality

(a 2 + 3a) 2 + 2(a 2 + 3a) = a(a + 1)(a + 2)(a + 3)

To solve the identity, we transform the left side of the equality by removing the common factor:

(a 2 + 3a) 2 + 2(a 2 + 3a) = (a 2 + 3a)(a 2 + 3a) + 2(a 2 + 3a) = (a 2 + 3a)(a 2 + 3a + 2 )

(a 2 + 3a)(a 2 + 3a + 2) = (a 2 + 3a)(a 2 + 2a + a + 2) = (a 2 + 3a)((a 2 + 2a) + (a + 2 ) = (a 2 + 3a)(a(a + 2) + (a + 2)) = (a 2 + 3a)(a + 1)(a + 2) = a(a + 3)(a + z )(a + 2) = a(a + 1)(a + 2)(a + 3)

Another example. Let us prove that for any values ​​of the variables x and y the expression

(x - y)(x + y) - 2x(x - y)

is not a positive number.

Solution. Let's take out the common factor x - y:

(x - y)(x + y) - 2x(x - y) = (x - y)(x + y - 2x) = (x - y)(y - x)

Let us note that we have obtained the product of two similar binomials, differing only in the order of the letters x and y. If we swapped the variables in one of the brackets, we would get the product of two identical expressions, that is, a square. But in order to swap x and y, you need to put a minus sign in front of the bracket:

(x - y) = -(y - x)

Then we can write:

(x - y)(y - x) = -(y - x)(y - x) = -(y - x) 2

As you know, the square of any number is greater than or equal to zero. This also applies to the expression (y - x) 2. If there is a minus in front of the expression, then it must be less than or equal to zero, that is, it is not a positive number.

Polynomial expansion helps solve some equations. The following statement is used:

If one part of the equation contains zero, and the other is a product of factors, then each of them should be equal to zero.

Example. Solve the equation (s - 1)(s + 1) = 0.

Solution. The product of the monomials s - 1 and s + 1 is written on the left side, and zero is written on the right side. Therefore, zero must equal either s - 1 or s + 1:

(s - 1)(s + 1) = 0

s - 1 = 0 or s + 1 = 0

s = 1 or s = -1

Each of the two obtained values ​​of the variable s is a root of the equation, that is, it has two roots.

Answer: -1; 1.

Example. Solve the equation 5w 2 - 15w = 0.

Solution. Let's take out 5w:

Again, the work is written on the left side, and a zero on the right. Let's continue with the solution:

5w = 0 or (w - 3) = 0

w = 0 or w = 3

Answer: 0; 3.

Example. Find the roots of the equation k 3 - 8k 2 + 3k- 24 = 0.

Solution. Let's group the terms:

k 3 - 8k 2 + 3k- 24 = 0

(k 3 - 8k 2) + (3k- 24) = 0

k 2 (k - 8) + 3(k - 8) = 0

(k 3 + 3)(k - 8) = 0

k 2 + 3 = 0 or k - 8 = 0

k 2 = -3 or k = 8

Note that the equation k 2 = - 3 has no solution, since any number squared is not less than zero. Therefore, the only root of the original equation is k = 8.

Example. Find the roots of the equation

(2u - 5)(u + 3) = 7u + 21

Solution: Move all terms to the left side, and then group the terms:

(2u - 5)(u + 3) = 7u + 21

(2u - 5)(u + 3) - 7u - 21 = 0

(2u - 5)(u + 3) - 7(u + 3) = 0

(2u - 5 - 7)(u + 3) = 0

(2u - 12)(u + 3) = 0

2u - 12 = 0 or u + 3 = 0

u = 6 or u = -3

Answer: - 3; 6.

Example. Solve the equation

(t 2 - 5t) 2 = 30t - 6t 2

(t 2 - 5t) 2 = 30t - 6t 2

(t 2 - 5t) 2 - (30t - 6t 2) = 0

(t 2 - 5t)(t 2 - 5t) + 6(t 2 - 5t) = 0

(t 2 - 5t)(t 2 - 5t + 6) = 0

t 2 - 5t = 0 or t 2 - 5t + 6 = 0

t = 0 or t - 5 = 0

t=0 or t=5

Now let's move on to the second equation. Again we have a quadratic trinomial. To factorize it using the grouping method, you need to present it as a sum of 4 terms. If you make the replacement - 5t = - 2t - 3t, then you can further group the terms:

t 2 - 5t + 6 = 0

t 2 - 2t - 3t + 6 = 0

t(t - 2) - 3(t - 2) = 0

(t - 3)(t - 2) = 0

T - 3 = 0 or t - 2 = 0

t=3 or t=2

As a result, we found that the original equation has 4 roots.

In general, this task requires a creative approach, since there is no universal method for solving it. But let's try to give a few tips.

In the overwhelming majority of cases, the factorization of a polynomial is based on a corollary of Bezout’s theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by . The root of the resulting polynomial is sought and the process is repeated until complete expansion.

If the root cannot be found, then specific expansion methods are used: from grouping to introducing additional mutually exclusive terms.

Further presentation is based on the skills of solving equations of higher degrees with integer coefficients.

Bracketing out the common factor.

Let's start with the simplest case, when the free term is equal to zero, that is, the polynomial has the form .

Obviously, the root of such a polynomial is , that is, we can represent the polynomial in the form .

This method is nothing more than putting the common factor out of brackets.

Example.

Factor a third degree polynomial.

Solution.

Obviously, what is the root of the polynomial, that is X can be taken out of brackets:

Let's find the roots of the quadratic trinomial

Thus,

Top of page

Factoring a polynomial with rational roots.

First, let's consider a method for expanding a polynomial with integer coefficients of the form , the coefficient of the highest degree is equal to one.

In this case, if a polynomial has integer roots, then they are divisors of the free term.

Example.

Solution.

Let's check if there are intact roots. To do this, write down the divisors of the number -18 : . That is, if a polynomial has integer roots, then they are among the written numbers. Let's check these numbers sequentially using Horner's scheme. Its convenience also lies in the fact that in the end we obtain the expansion coefficients of the polynomial:

That is, x=2 And x=-3 are the roots of the original polynomial and we can represent it as a product:

It remains to expand the quadratic trinomial.

The discriminant of this trinomial is negative, therefore it has no real roots.

Answer:

Comment:

Instead of Horner's scheme, one could use the selection of the root and subsequent division of the polynomial by a polynomial.

Now consider the expansion of a polynomial with integer coefficients of the form , and the coefficient of the highest degree is not equal to one.

In this case, the polynomial can have fractionally rational roots.

Example.

Factor the expression.

Solution.

By performing a variable change y=2x, let's move on to a polynomial with a coefficient equal to one at the highest degree. To do this, first multiply the expression by 4 .

If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down:

Let us sequentially calculate the values ​​of the function g(y) at these points until zero is reached.

That is, y=-5 is the root , therefore, is the root of the original function. Let's divide the polynomial by a column (corner) into a binomial.

Thus,

It is not advisable to continue checking the remaining divisors, since it is easier to factorize the resulting quadratic trinomial

Hence,

Unknown polynomials. The theorem about the distribution of polynomials in additions of unknowns. Canonical layout of a polynomial.

8 examples of factoring polynomials are given. They include examples of solving quadratic and biquadratic equations, examples of reciprocal polynomials, and examples of finding integer roots of third- and fourth-degree polynomials.

Content

See also: Methods for factoring polynomials
Roots of a quadratic equation
Solving cubic equations

1. Examples with solving a quadratic equation

Example 1.1

x 4 + x 3 - 6 x 2.

We take out x 2 outside of brackets:
.
2 + x - 6 = 0:
.
Roots of the equation:
, .

.

Example 1.2

Factor the third degree polynomial:
x 3 + 6 x 2 + 9 x.

Let's take x out of brackets:
.
Solving the quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant: .
Since the discriminant is zero, the roots of the equation are multiples: ;
.

From here we obtain the factorization of the polynomial:
.

Example 1.3

Factor the fifth degree polynomial:
x 5 - 2 x 4 + 10 x 3.

We take out x 3 outside of brackets:
.
Solving the quadratic equation x 2 - 2 x + 10 = 0.
Its discriminant: .
Since the discriminant is less than zero, the roots of the equation are complex: ;
, .

The factorization of the polynomial has the form:
.

If we are interested in factorization with real coefficients, then:
.

Examples of factoring polynomials using formulas

Examples with biquadratic polynomials

Example 2.1

Factor the biquadratic polynomial:
x 4 + x 2 - 20.

Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).

;
.

Example 2.2

Factor the polynomial that reduces to a biquadratic one:
x 8 + x 4 + 1.

Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):

;

;
.

Example 2.3 with recurrent polynomial

Factor the reciprocal polynomial:
.

A reciprocal polynomial has odd degree. Therefore it has root x = - 1 . Divide the polynomial by x - (-1) = x + 1. As a result we get:
.
Let's make a substitution:
, ;
;


;
.

Examples of factoring polynomials with integer roots

Example 3.1

Factor the polynomial:
.

Let's assume that 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 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 found three roots, they are simple. Then
.

Example 3.2

Factor the polynomial:
.

Let's assume that the equation

has at least one whole root. Then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
-2, -1, 1, 2 .
We 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.

So, we found one root:
x 1 = -1 .
Divide the polynomial by x - x 1 = x - (-1) = x + 1:


Then,
.

Now we need to solve the third degree equation:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Let's 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:
.

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 polynomial expansion 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, so 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, 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 .

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 a common value is given 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 decomposing 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, let us 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.

The article will cover all the concepts of decomposition, theoretical foundations and methods of factoring a polynomial.

Theory

Theorem 1

When 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 2

Any 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).

This shows that the expansion itself reduces to solving the 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 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 = 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

Expand 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, you need 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.

If you notice an error in the text, please highlight it and press Ctrl+Enter