How to decompose a square trinomial into linear factors. Factoring a polynomial

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

See also: Methods for factoring polynomials
The roots of a quadratic equation
Solution of cubic equations

1. Examples with the solution of a quadratic equation

Example 1.1

x 4 + x 3 - 6 x 2.

Take out x 2 for brackets:
.
2 + x - 6 = 0:
.
Equation roots:
, .

.

Example 1.2

Factoring a third-degree polynomial:
x 3 + 6 x 2 + 9 x.

We take x out of brackets:
.
We solve the quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant is .
Since the discriminant is equal to zero, the roots of the equation are multiples: ;
.

From here we obtain the decomposition of the polynomial into factors:
.

Example 1.3

Factoring a fifth-degree polynomial:
x 5 - 2 x 4 + 10 x 3.

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:
.

Examples of factoring polynomials using formulas

Examples with biquadratic polynomials

Example 2.1

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

Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).

;
.

Example 2.2

Factoring a polynomial that reduces to a biquadratic:
x 8 + x 4 + 1.

Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):

;

;
.

Example 2.3 with recursive polynomial

Factoring the recursive polynomial:
.

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:
, ;
;


;
.

Examples of Factoring Polynomials with Integer Roots

Example 3.1

Factoring a polynomial:
.

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
.

Example 3.2

Factoring a polynomial:
.

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.

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


Then,
.

Now we need to solve the equation of the third degree:
.
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:
.

He has a square, and it consists of three terms (). So it turns out - a square trinomial.

Examples not square trinomials:

\(x^3-3x^2-5x+6\) - cubic quaternary
\(2x+1\) - linear binomial

The root of the square trinomial:

Example:
The trinomial \(x^2-2x+1\) has a root \(1\), because \(1^2-2 1+1=0\)
The trinomial \(x^2+2x-3\) has roots \(1\) and \(-3\), because \(1^2+2-3=0\) and \((-3)^ 2-6-3=9-9=0\)

For example: if you need to find the roots for the square trinomial \(x^2-2x+1\), we equate it to zero and solve the equation \(x^2-2x+1=0\).

\(D=4-4\cdot1=0\)
\(x=\frac(2-0)(2)=\frac(2)(2)=1\)

Ready. The root is \(1\).

Decomposition of a square trinomial into:

The square trinomial \(ax^2+bx+c\) can be expanded as \(a(x-x_1)(x-x_2)\) if the equations \(ax^2+bx+c=0\) are greater than zero \ (x_1\) and \(x_2\) are the roots of the same equation).

For example, consider the trinomial \(3x^2+13x-10\).
The quadratic equation \(3x^2+13x-10=0\) has a discriminant equal to 289 (greater than zero), and the roots are equal to \(-5\) and \(\frac(2)(3)\). So \(3x^2+13x-10=3(x+5)(x-\frac(2)(3))\). It is easy to verify the correctness of this statement - if we , then we get the original trinomial.

The square trinomial \(ax^2+bx+c\) can be represented as \(a(x-x_1)^2\) if the discriminant of the equation \(ax^2+bx+c=0\) is equal to zero.

The square trinomial \(ax^2+bx+c\) does not factorize if the discriminant of the equation \(ax^2+bx+c=0\) is less than zero.

For example, the trinomials \(x^2+x+4\) and \(-5x^2+2x-1\) have a discriminant less than zero. Therefore, it is impossible to decompose them into factors.

Example . Factor \(2x^2-11x+12\).
Solution :
Find the roots of the quadratic equation \(2x^2-11x+12=0\)

\(D=11^2-4 \cdot 2 \cdot 12=121-96=25>0\)
\(x_1=\frac(11-5)(4)=1.5;\) \(x_2=\frac(11+5)(4)=4.\)

So \(2x^2-11x+12=2(x-1,5)(x-4)\)
Answer : \(2(x-1.5)(x-4)\)

The received answer may be written in a different way: \((2x-3)(x-4)\).

Example . (Assignment from the OGE) The square trinomial is factored \(5x^2+33x+40=5(x++ 5)(x-a)\). Find \(a\).
Solution:
\(5x^2+33x+40=0\)
\(D=33^2-4 \cdot 5 \cdot 40=1089-800=289=17^2\)
\(x_1=\frac(-33-17)(10)=-5\)
\(x_2=\frac(-33+17)(10)=-1.6\)
\(5x^2+33x+40=5(x+5)(x+1,6)\)
Answer : \(-1,6\)

SQUARE TRIPON III

§ 54. Decomposition of a square trinomial into linear factors

In this section, we consider the following question: in which case the square trinomial ax 2 + bx+c can be represented as a product

(a 1 x+b 1) (a 2 x+b 2)

two linear relatively X factors with real coefficients a 1 , b 1 , a 2 , b 2 (a 1 =/=0, a 2 =/=0) ?

1. Assume that the given square trinomial ax 2 + bx+c represent in the form

ax 2 + bx+c = (a 1 x+b 1) (a 2 x+b 2). (1)

The right side of formula (1) vanishes when X = - b 1 / a 1 and X = - b 2 / a 2 (a 1 and a 2 are not equal to zero by condition). But in this case, the numbers b 1 / a 1 and - b 2 / a 2 are the roots of the equation

ax 2 + bx+c = 0.

Therefore, the discriminant of the square trinomial ax 2 + bx+c must be non-negative.

2. Conversely, suppose that the discriminant D = b 2 - 4ace square trinomial ax 2 + bx+c is non-negative. Then this trinomial has real roots x 1 and x 2. Using the Vieta theorem, we get:

ax 2 + bx+c =a (x 2 + b / a X + c / a ) = a [x 2 - (x 1 + x 2) X + x 1 x 2 ] =

= a [(x 2 - x 1 x ) - (x 2 x - x 1 x 2)] = a [X (X - x 1) - x 2 (X - x 1) =

=a (X - x 1)(X - x 2).

ax 2 + bx+c = a (X - x 1)(X - x 2), (2)

where x 1 and x 2 - roots of the trinomial ax 2 + bx+c . Coefficient a can be attributed to either of two linear factors, for example,

a (X - x 1)(X - x 2) = (ah - ax 1)(X - x 2).

But this means that in the case under consideration the square trinomial ax 2 + bx+c represent as a product of two linear factors with real coefficients.

Combining the results obtained in sections 1 and 2, we arrive at the following theorem.

Theorem. Square trinomial ax 2 + bx+c then and only then can be represented as a product of two linear factors with real coefficients,

ax 2 + bx+c = (ah - ax 1)(X - x 2),

when the discriminant of this square trinomial is non-negative (that is, when this trinomial has real roots).

Example 1. Factorize into linear factors 6 x 2 - X -1.

The roots of this square trinomial are x 1 = 1 / 2 and x 2 = - 1 / 3 .

Therefore, according to formula (2)

6x 2 - X -1 = 6 (X - 1 / 2)(X + 1 / 3) = (2X - 1) (3x + 1).

Example 2. Factorize into Linear Factors x 2 + X + 1. The discriminant of this square trinomial is negative:

D = 1 2 - 4 1 1 = - 3< 0.

Therefore, this square trinomial cannot be decomposed into linear factors with real coefficients.

Exercises

Expand the following expressions into linear factors (No. 403 - 406):

403. 6x 2 - 7X + 2. 405. x 2 - X + 1.

404. 2x 2 - 7Oh + 6a 2 . 406. x 2 - 3Oh + 2a 2 - ab - b 2 .

Reduce fractions (No. 407, 408):

Solve Equations:

The square trinomial can be factored as follows:

A x 2 + b x + c = a ⋅ (x − x 1) ⋅ (x − x 2)

where a is the number, coefficient before the highest coefficient,

x is a variable (that is, a letter),

x 1 and x 2 - numbers, roots of the quadratic equation a x 2 + b x + c \u003d 0, which are found through the discriminant.

If the quadratic equation has only one root, then the decomposition looks like this:

a x 2 + b x + c = a ⋅ (x − x 0) 2

Examples of factoring a square trinomial:

1. − x 2 + 6 x + 7 = 0 ⇒ x 1 = − 1,   x 2 = 7

− x 2 + 6 x + 7 = (− 1) ⋅ (x − (− 1)) (x − 7) = − (x + 1) (x − 7) = (x + 1) (7 − x)

1. − x 2 + 4 x − 4 = 0; ⇒ x0 = 2

− x 2 + 4 x − 4 = (− 1) ⋅ (x − 2) 2 = − (x − 2) 2

If a square trinomial is incomplete (b = 0 or c = 0), then it can be factored in the following ways:

• c = 0 ⇒ a x 2 + b x = x (a x + b)

• b = 0 ⇒ apply the reduced multiplication formula for the difference of squares.

Tasks for independent solution

No. 1. The square trinomial is factorized: x 2 + 6 x - 27 = (x + 9) (x - a) . Find a .

Solution:

First you need to equate the square trinomial to zero to find x 1 and x 2.

x 2 + 6 x − 27 = 0

a = 1, b = 6, c = − 27

D = b 2 − 4 a c = 6 2 − 4 ⋅ 1 ⋅ (− 27) = 36 + 108 = 144

D > 0 means there will be two different roots.

x 1,2 = − b ± D 2 a = − 6 ± 144 2 ⋅ 1 = [ − 6 + 12 2 = 6 2 = 3 − 6 − 12 2 = − 18 2 = − 9

Knowing the roots, we factorize the square trinomial:

x 2 + 6 x − 27 = (x − (− 9)) (x − 3) = (x + 9) (x − 3)

No. 2. The equation x 2 + p x + q \u003d 0 has roots - 5; 7. Find q.

Solution:

1 way:(you need to know how the square trinomial is factored)

If x 1 and x 2 are the roots of a square trinomial a x 2 + b x + c, then it can be factored as follows: a x 2 + b x + c = a ⋅ (x − x 1) ⋅ (x − x 2) .

Since in a given square trinomial the leading coefficient (the factor in front of x 2) is equal to one, the decomposition will be as follows:

x 2 + p x + q = (x − x 1) (x − x 2) = (x − (− 5)) (x − 7) = (x + 5) (x − 7) = x 2 − 7 x + 5 x - 35 = x 2 - 2 x - 35

x 2 + p x + q = x 2 − 2 x − 35 ⇒ p = − 2, q = − 35

2 way: (you need to know the Vieta theorem)

Vieta's theorem:

The sum of the roots of the reduced square trinomial x 2 + p x + q is equal to its second coefficient p with the opposite sign, and the product is equal to the free term q.

( x 1 + x 2 = − p x 1 ⋅ x 2 = q

q = x 1 ⋅ x 2 = (− 5) ⋅ 7 = − 35.

It is necessary to factorize polynomials when simplifying expressions (so that reduction can be made), when solving equations, or when decomposing a fractionally rational function into simple fractions.

It makes sense to talk about factoring a polynomial if its degree is not lower than the second.

The polynomial of the first degree is called linear.

Let us first consider the theoretical foundations, then proceed directly to the methods of factoring a polynomial.

Page navigation.

Necessary theory.

Theorem.

Any degree polynomial n of the form is represented by the product of a constant factor at the highest degree and n linear multipliers , i=1, 2, …, n, that is, , and , i=1, 2, …, n are the roots of the polynomial.

This theorem is formulated for complex roots, i=1, 2, …, n and complex coefficients , k=0, 1, 2, …, n. It is the basis for factoring any polynomial.

If the coefficients k=0, 1, 2, …, n are real numbers, then the complex roots of the polynomial will MANDATORY occur in complex conjugate pairs.

For example, if the roots and the polynomial are complex conjugate, and the remaining roots are real, then the polynomial will be represented as , where

Comment.

Among the roots of a polynomial, there may be repeating ones.

The proof of the theorem is carried out using fundamental theorem of algebra and corollaries from Bezout's theorem.

Fundamental theorem of algebra.

Any polynomial of degree n has at least one root (complex or real).

Bezout's theorem.

When dividing a polynomial by (x-s) the remainder is equal to the value of the polynomial at the point s, i.e. , where is a polynomial of degree n-1.

Corollary from Bezout's theorem.

If a s is the root of the polynomial , then .

We will often use this corollary when describing the solution of examples.

Factorization of a square trinomial.

The square trinomial is decomposed into two linear factors: , where and are roots (complex or real).

Thus, factoring a quadratic trinomial reduces to solving a quadratic equation.

Example.

Factorize the square trinomial.

Solution.

Find the roots of the quadratic equation .

The discriminant of the equation is , therefore,

In this way, .

To check, you can open the brackets: . When checking, we came to the original trinomial, so the expansion is correct.

Example.

Solution.

The corresponding quadratic equation has the form .

Let's find its roots.

That's why, .

Example.

Factorize the polynomial.

Solution.

Let's find the roots of the quadratic equation.

Get a pair of complex conjugate roots.

The expansion of the polynomial will have the form .

Example.

Factorize the square trinomial.

Solution.

Let's solve the quadratic equation .

That's why,

Comment:

In the future, with a negative discriminant, we will leave the second-order polynomials in their original form, that is, we will not decompose them into linear factors with complex free terms.

Methods for factoring a polynomial of degree higher than the second.

In the general case, this task involves a creative approach, since there is no universal method for solving it. However, let's try to give a few hints.

In the vast majority of cases, the decomposition of a polynomial into factors is based on the consequence of the Bezout theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by. The resulting polynomial is searched for a root and the process is repeated until complete expansion.

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

What follows is based on skills with integer coefficients.

Bracketing 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, the polynomial can be represented as .

This method is nothing but taking the common factor out of brackets.

Example.

Decompose a polynomial of the third degree into factors.

Solution.

It is obvious that is the root of the polynomial, that is, X can be bracketed:

Find the roots of a square trinomial

In this way,

Factorization of a polynomial with rational roots.

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

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

Example.

Solution.

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

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

It remains to expand the square trinomial.

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

Answer:

Comment:

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

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

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

Example.

Factorize the expression.

Solution.

By changing the variable y=2x, we pass to a polynomial with a coefficient equal to one at the highest degree. To do this, we 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:

Calculate sequentially the values ​​of the function g(y) at these points until reaching zero.

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

In this way,

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

Consequently,

Artificial tricks in the decomposition of a polynomial into factors.

Polynomials do not always have rational roots. In this case, when factoring, one has to look for special methods. But, no matter how much we would like, some polynomials (or rather, the vast majority) cannot be represented as a product.

grouping method.

Sometimes it turns out to group the terms of a polynomial, which allows you to find a common factor and take it out of brackets.

Example.

Expand polynomial for multipliers.

Solution.

Since the coefficients are integers, there can be integer roots among the divisors of the free term. Let's check the values 1 , -1 , 2 and -2 , calculating the value of the polynomial at these points.

That is, there are no whole roots. We will look for another way of decomposition.

Let's group:

After grouping, the original polynomial was presented as a product of two square trinomials. Let's factor them out.