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Factorization of square trinomials: examples and formulas. Square trinomial

The factorization of square trinomials is one of the school assignments that everyone faces sooner or later. How to do it? What is the formula for factoring a square trinomial? Let's go through it step by step with examples.

General formula

The factorization of square trinomials is carried out by solving a quadratic equation. This is a simple task that can be solved by several methods - by finding the discriminant, using the Vieta theorem, there is also a graphical way to solve it. The first two methods are studied in high school.

The general formula looks like this:lx 2 +kx+n=l(x-x 1)(x-x 2) (1)

Task execution algorithm

In order to factorize square trinomials, you need to know Wit's theorem, have a program for solving at hand, be able to find a solution graphically or look for the roots of a second-degree equation through the discriminant formula. If a square trinomial is given and it must be factored, the algorithm of actions is as follows:

1) Equate the original expression to zero to get the equation.

2) Give similar terms (if necessary).

3) Find the roots by any known method. The graphical method is best used if it is known in advance that the roots are integers and small numbers. It must be remembered that the number of roots is equal to the maximum degree of the equation, that is, the quadratic equation has two roots.

4) Substitute value X into expression (1).

5) Write down the factorization of square trinomials.

Examples

Practice allows you to finally understand how this task is performed. Examples illustrate the factorization of a square trinomial:

you need to expand the expression:

Let's use our algorithm:

1) x 2 -17x+32=0

2) similar terms are reduced

3) according to the Vieta formula, it is difficult to find the roots for this example, therefore it is better to use the expression for the discriminant:

D=289-128=161=(12.69) 2

4) Substitute the roots we found in the main formula for expansion:

(x-2.155) * (x-14.845)

5) Then the answer will be:

x 2 -17x + 32 \u003d (x-2.155) (x-14.845)

Let's check whether the solutions found by the discriminant correspond to the Vieta formulas:

14,845 . 2,155=32

For these roots, Vieta's theorem is applied, they were found correctly, which means that the factorization we obtained is also correct.

Similarly, we expand 12x 2 + 7x-6.

x 1 \u003d -7 + (337) 1/2

x 2 \u003d -7- (337) 1/2

In the previous case, the solutions were non-integer, but real numbers, which are easy to find with a calculator in front of you. Now consider a more complex example in which the roots are complex: factorize x 2 + 4x + 9. According to the Vieta formula, the roots cannot be found, and the discriminant is negative. The roots will be on the complex plane.

D=-20

Based on this, we get the roots we are interested in -4 + 2i * 5 1/2 and -4-2i * 5 1/2 because (-20) 1/2 = 2i*5 1/2 .

We obtain the desired expansion by substituting the roots into the general formula.

Another example: you need to factorize the expression 23x 2 -14x + 7.

We have the equation 23x 2 -14x+7 =0

D=-448

So the roots are 14+21,166i and 14-21,166i. The answer will be:

23x 2 -14x+7 =23(x- 14-21,166i )*(X- 14+21.166i ).

Let us give an example that can be solved without the help of the discriminant.

Let it be necessary to decompose the quadratic equation x 2 -32x + 255. Obviously, it can also be solved by the discriminant, but it is faster in this case to find the roots.

x 1 =15

x2=17

Means x 2 -32x + 255 =(x-15)(x-17).

In this lesson, we will learn how to decompose square trinomials into linear factors. For this, it is necessary to recall Vieta's theorem and its inverse. This skill will help us quickly and conveniently decompose square trinomials into linear factors, and also simplify the reduction of fractions consisting of expressions.

So back to the quadratic equation , where .

What we have on the left side is called the square trinomial.

The theorem is true: If are the roots of a square trinomial, then the identity is true

Where is the leading coefficient, are the roots of the equation.

So, we have a quadratic equation - a square trinomial, where the roots of the quadratic equation are also called the roots of the quadratic trinomial. Therefore, if we have the roots of a square trinomial, then this trinomial is decomposed into linear factors.

Proof:

The proof of this fact is carried out using the Vieta theorem, which we considered in previous lessons.

Let's remember what Vieta's theorem tells us:

If are the roots of a square trinomial for which , then .

This theorem implies the following assertion that .

We see that, according to the Vieta theorem, i.e., substituting these values into the formula above, we get the following expression

Q.E.D.

Recall that we proved the theorem that if are the roots of a square trinomial, then the decomposition is valid.

Now let's recall an example of a quadratic equation, to which we selected the roots using Vieta's theorem. From this fact we can obtain the following equality thanks to the proved theorem:

Now let's check the correctness of this fact by simply expanding the brackets:

We see that we factored correctly, and any trinomial, if it has roots, can be factored according to this theorem into linear factors according to the formula

However, let's check whether for any equation such a factorization is possible:

Let's take the equation for example. First, let's check the sign of the discriminant

And we remember that in order to fulfill the theorem we have learned, D must be greater than 0, therefore, in this case, factoring according to the studied theorem is impossible.

Therefore, we formulate a new theorem: if a square trinomial has no roots, then it cannot be decomposed into linear factors.

So, we have considered the Vieta theorem, the possibility of decomposing a square trinomial into linear factors, and now we will solve several problems.

Task #1

In this group, we will actually solve the problem inverse to the one posed. We had an equation, and we found its roots, decomposing into factors. Here we will do the opposite. Let's say we have the roots of a quadratic equation

The inverse problem is this: write a quadratic equation so that were its roots.

There are 2 ways to solve this problem.

Since are the roots of the equation, then is a quadratic equation whose roots are given numbers. Now let's open the brackets and check:

This was the first way we created a quadratic equation with given roots that does not have any other roots, since any quadratic equation has at most two roots.

This method involves the use of the inverse Vieta theorem.

If are the roots of the equation, then they satisfy the condition that .

For the reduced quadratic equation , , i.e. in this case , and .

Thus, we have created a quadratic equation that has the given roots.

Task #2

You need to reduce the fraction.

We have a trinomial in the numerator and a trinomial in the denominator, and the trinomials may or may not be factorized. If both the numerator and the denominator are factorized, then among them there may be equal factors that can be reduced.

First of all, it is necessary to factorize the numerator.

First, you need to check whether this equation can be factored, find the discriminant . Since , then the sign depends on the product ( must be less than 0), in this example , i.e., the given equation has roots.

To solve, we use the Vieta theorem:

In this case, since we are dealing with roots, it will be quite difficult to simply pick up the roots. But we see that the coefficients are balanced, i.e. if we assume that , and substitute this value into the equation, then the following system is obtained: i.e. 5-5=0. Thus, we have chosen one of the roots of this quadratic equation.

We will look for the second root by substituting what is already known into the system of equations, for example, , i.e. .

Thus, we have found both roots of the quadratic equation and can substitute their values into the original equation to factor it:

Recall the original problem, we needed to reduce the fraction.

Let's try to solve the problem by substituting instead of the numerator .

It is necessary not to forget that in this case the denominator cannot be equal to 0, i.e.,.

If these conditions are met, then we have reduced the original fraction to the form .

Task #3 (task with a parameter)

At what values of the parameter is the sum of the roots of the quadratic equation

If the roots of this equation exist, then , the question is when .

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.

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 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:
.

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.

A square trinomial is a polynomial of the form ax^2+bx+c, where x is a variable, a, b and c are some numbers, and a is not equal to zero.
Actually, the first thing we need to know in order to factorize the ill-fated trinomial is the theorem. It looks like this: “If x1 and x2 are the roots of the square trinomial ax^2+bx+c, then ax^2+bx+c=a(x-x1)(x-x2)”. Of course, there is also a proof of this theorem, but it requires some theoretical knowledge (if we take out the factor a in the polynomial ax^2+bx+c we get ax^2+bx+c=a(x^2+(b/a) x + c/a) By Viette's theorem x1+x2=-(b/a), x1*x2=c/a, hence b/a=-(x1+x2), c/a=x1*x2. , x^2+ (b/a)x+c/a= x^2- (x1+x2)x+ x1x2=x^2-x1x-x2x+x1x2=x(x-x1)-x2(x-x1 )= (x-x1)(x-x2), so ax^2+bx+c=a(x-x1)(x-x2) Sometimes teachers make you learn the proof, but if it is not required, I advise you to just remember final formula.

2 step

Let's take as an example the trinomial 3x^2-24x+21. The first thing we need to do is equate the trinomial to zero: 3x^2-24x+21=0. The roots of the resulting quadratic equation will be the roots of the trinomial, respectively.

3 step

Solve the equation 3x^2-24x+21=0. a=3, b=-24, c=21. So, let's decide. Who does not know how to solve quadratic equations, look at my instructions with 2 ways to solve them using the example of the same equation. We got the roots x1=7, x2=1.

4 step

Now that we have the trinomial roots, we can safely substitute them into the formula =) ax^2+bx+c=a(x-x1)(x-x2)
we get: 3x^2-24x+21=3(x-7)(x-1)
You can get rid of the term a by putting it in brackets: 3x^2-24x+21=(x-7)(x*3-1*3)
as a result we get: 3x^2-24x+21=(x-7)(3x-3). Note: each of the obtained factors ((x-7), (3x-3) are polynomials of the first degree. That's the whole expansion =) If you doubt the answer you got, you can always check it by multiplying the brackets.

5 step

Verification of the solution. 3x^2-24x+21=3(x-7)(x-3)
(x-7)(3x-3)=3x^2-3x-21x+21=3x^2-24x+21. Now we know for sure that our solution is correct! I hope my instructions help someone =) Good luck with your studies!

In our case, in the equation D > 0 and we got 2 roots each. If it were D<0, то уравнение, как и многочлен, соответственно, корней бы не имело.
If a square trinomial has no roots, then it cannot be factored into factors that are polynomials of the first degree.