Decompose a quadratic trinomial into a binomial. Factoring a quadratic trinomial

Factoring quadratic trinomials is one of the school assignments that everyone faces sooner or later. How to do it? What is the formula for factoring a quadratic trinomial? Let's figure it out step by step using examples.

General formula

Quadratic trinomials are factorized by solving a quadratic equation. This is a simple problem that can be solved by several methods - by finding the discriminant, using Vieta's theorem, there is also a graphical solution. 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)

Algorithm for completing the task

In order to factor quadratic trinomials, you need to know Vita's theorem, have a solution program at hand, be able to find a solution graphically, or look for roots of a second-degree equation using the discriminant formula. If a quadratic trinomial is given and it needs to be factorized, the algorithm is as follows:

1) Equate the original expression to zero to obtain an equation.

2) Give similar terms (if necessary).

3) Find the roots using 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 the value X into expression (1).

5) Write down the factorization of quadratic trinomials.

Examples

Practice allows you to finally understand how this task is performed. The following examples illustrate the factorization of a quadratic trinomial:

it is necessary to expand the expression:

Let's resort to our algorithm:

1) x 2 -17x+32=0

2) similar terms are reduced

3) using Vieta’s formula, it is difficult to find roots for this example, so it is better to use the expression for the discriminant:

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

4) Let’s substitute the roots we found into the basic formula for decomposition:

(x-2.155) * (x-14.845)

5) Then the answer will be like this:

x 2 -17x+32=(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 the factorization we obtained is also correct.

Let us similarly expand 12x 2 + 7x-6.

x 1 =-7+(337) 1/2

x 2 =-7-(337)1/2

In the previous case, the solutions were non-integer, but real numbers, which are easy to find if you have a calculator in front of you. Now let's look at a more complex example, in which the roots will be complex: factor x 2 + 4x + 9. Using Vieta's 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 obtain the roots that interest us -4+2i*5 1/2 and -4-2i * 5 1/2 since (-20) 1/2 = 2i*5 1/2 .

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

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

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

D=-448

This means 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 a discriminant.

Let's say we need to expand the quadratic equation x 2 -32x+255. Obviously, it can also be solved using a discriminant, but in this case it is faster to find the roots.

x 1 =15

x 2 =17

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

This online calculator is designed to factorize a function.

For example, factorize: x 2 /3-3x+12. Let's write it as x^2/3-3*x+12. You can also use this service, where all calculations are saved in Word format.

For example, decompose into terms. Let's write it as (1-x^2)/(x^3+x) . To see the progress of the solution, click Show steps. If you need to get the result in Word format, use this service.

Note: the number "pi" (π) is written as pi; square root as sqrt , for example sqrt(3) , tangent tg is written tan . To view the answer, see Alternative.

1. If a simple expression is given, for example, 8*d+12*c*d, then factoring the expression means representing the expression in the form of factors. To do this, you need to find common factors. Let's write this expression as: 4*d*(2+3*c) .
2. Present the product in the form of two binomials: x 2 + 21yz + 7xz + 3xy. Here you already need to find several common factors: x(x+7z) + 3y(x + 7z). We take out (x+7z) and get: (x+7z)(x + 3y) .

see also Division of polynomials with a corner (all steps of division with a column are shown)

Useful when studying the rules of factorization will be abbreviated multiplication formulas, with the help of which it will be clear how to open brackets with a square:

1. (a+b) 2 = (a+b)(a+b) = a 2 +2ab+b 2
2. (a-b) 2 = (a-b)(a-b) = a 2 -2ab+b 2
3. (a+b)(a-b) = a 2 - b 2
4. a 3 +b 3 = (a+b)(a 2 -ab+b 2)
5. a 3 -b 3 = (a-b)(a 2 +ab+b 2)
6. (a+b) 3 = (a+b)(a+b) 2 = a 3 +3a 2 b + 3ab 2 +b 3
7. (a-b) 3 = (a-b)(a-b) 2 = a 3 -3a 2 b + 3ab 2 -b 3

Factorization Methods

After learning a few techniques factorization The following classification of solutions can be made:

1. Using abbreviated multiplication formulas.
2. Finding a common factor.

In this lesson we will learn how to factor quadratic trinomials into linear factors. To do this, we need to remember Vieta’s theorem and its converse. This skill will help us quickly and conveniently expand quadratic trinomials into linear factors, and will also simplify the reduction of fractions consisting of expressions.

So let's go back to the quadratic equation, where .

What we have on the left side is called a quadratic trinomial.

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

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

So, we have a quadratic equation - a quadratic 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 can be decomposed into linear factors.

Proof:

The proof of this fact is carried out using Vieta’s theorem, which we discussed in previous lessons.

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

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

The following statement follows from this theorem:

We see that, according to Vieta’s theorem, i.e., by substituting these values ​​into the formula above, we obtain the following expression

Q.E.D.

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

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

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

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

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

Take, for example, the equation . First, let's check the discriminant sign

And we remember that in order to fulfill the theorem we learned, D must be greater than 0, so in this case, factorization according to the theorem we learned 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 looked at Vieta’s theorem, the possibility of decomposing a quadratic trinomial into linear factors, and now we will solve several problems.

Task No. 1

In this group we will actually solve the problem inverse to the one posed. We had an equation, and we found its roots by factoring it. 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 using 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 in which we created a quadratic equation with given roots, which 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 No. 2

It is necessary 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 factored, then among them there may be equal factors that can be reduced.

First of all, you need to factor the numerator.

First, you need to check whether this equation can be factorized, let's find the discriminant. Since , 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 Vieta’s theorem:

In this case, since we are dealing with roots, it will be quite difficult to simply select the roots. But we see that the coefficients are balanced, that is, if we assume that , and substitute this value into the equation, we get the following system: , i.e. 5-5=0. Thus, we have selected 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:

Let's remember the original problem, we needed to reduce the fraction .

Let's try to solve the problem by substituting .

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 .

Problem No. 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 , question: when.

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 factor 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 a proof of this theorem, but it requires some theoretical knowledge (when 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, therefore 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). This means ax^2+bx+c=a(x-x1)(x-x2) Sometimes teachers force you to learn the proof, but if it is not required, I advise you to just memorize it. final formula.

Step 2

Let's take the trinomial 3x^2-24x+21 as an example. 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.

Step 3

Let's solve the equation 3x^2-24x+21=0. a=3, b=-24, c=21. So, let's decide. For those who don’t know how to solve quadratic equations, look at my instructions with 2 ways to solve them using the same equation as an example. The resulting roots are x1=7, x2=1.

Step 4

Now that we have the roots of the trinomial, 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 a term 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 resulting factors ((x-7), (3x-3) are polynomials of the first degree. That’s all the expansion =) If you doubt the answer received, you can always check it by multiplying the brackets.

Step 5

Checking 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 decision is correct! I hope my instructions will help someone =) Good luck with your studies!

• In our case, in the equation D > 0 and we got 2 roots. If there was a D<0, то уравнение, как и многочлен, соответственно, корней бы не имело.
• If a square trinomial has no roots, then it cannot be factorized, which are polynomials of the first degree.

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 ≠ 0.

To factor a trinomial, you need to know the roots of that trinomial. (further an example on the trinomial 5x^2 + 3x- 2)

Note: the value of the quadratic trinomial 5x^2 + 3x - 2 depends on the value of x. For example: If x = 0, then 5x^2 + 3x - 2 = -2

If x = 2, then 5x^2 + 3x - 2 = 24

If x = -1, then 5x^2 + 3x - 2 = 0

At x = -1, the square trinomial 5x^2 + 3x - 2 vanishes, in this case the number -1 is called root of a square trinomial.

How to get the root of an equation

Let us explain how we obtained the root of this equation. First, you need to clearly know the theorem and the formula by which we will work:

“If x1 and x2 are the roots of the quadratic trinomial ax^2 + bx + c, then ax^2 + bx + c = a(x - x1)(x - x2).”

X = (-b±√(b^2-4ac))/2a\

This formula for finding the roots of a polynomial is the most primitive formula, using which you will never get confused.

The expression is 5x^2 + 3x – 2.

1. Equate to zero: 5x^2 + 3x – 2 = 0

2. Find the roots of the quadratic equation, to do this we substitute the values ​​into the formula (a is the coefficient of X^2, b is the coefficient of X, the free term, that is, the figure without X):

We find the first root with a plus sign in front of the square root:

Х1 = (-3 + √(3^2 - 4 * 5 * (-2)))/(2*5) = (-3 + √(9 -(-40)))/10 = (-3 + √(9+40))/10 = (-3 + √49)/10 = (-3 +7)/10 = 4/(10) = 0.4

The second root with a minus sign in front of the square root:

X2 = (-3 - √(3^2 - 4 * 5 * (-2)))/(2*5) = (-3 - √(9- (-40)))/10 = (-3 - √(9+40))/10 = (-3 - √49)/10 = (-3 - 7)/10 = (-10)/(10) = -1

So we have found the roots of the quadratic trinomial. To make sure that they are correct, you can check: first we substitute the first root into the equation, then the second:

1) 5x^2 + 3x – 2 = 0

5 * 0,4^2 + 3*0,4 – 2 = 0

5 * 0,16 + 1,2 – 2 = 0

2) 5x^2 + 3x – 2 = 0

5 * (-1)^2 + 3 * (-1) – 2 = 0

5 * 1 + (-3) – 2 = 0

5 – 3 – 2 = 0

If, after substituting all the roots, the equation becomes zero, then the equation is solved correctly.

3. Now let’s use the formula from the theorem: ax^2 + bx + c = a(x-x1)(x-x2), remember that X1 and X2 are the roots of the quadratic equation. So: 5x^2 + 3x – 2 = 5 * (x - 0.4) * (x- (-1))

5x^2 + 3x– 2 = 5(x - 0.4)(x + 1)

4. To make sure that the decomposition is correct, you can simply multiply the brackets:

5(x - 0.4)(x + 1) = 5(x^2 + x - 0.4x - 0.4) = 5(x^2 + 0.6x – 0.4) = 5x^2 + 3 – 2. Which confirms the correctness of the decision.

The second option for finding the roots of a square trinomial

Another option for finding the roots of a square trinomial is the inverse theorem to Viette’s theorem. Here the roots of the quadratic equation are found using the formulas: x1 + x2 = -(b), x1 * x2 = c. But it is important to understand that this theorem can only be used if the coefficient a = 1, that is, the number in front of x^2 = 1.

For example: x^2 – 2x +1 = 0, a = 1, b = - 2, c = 1.

We solve: x1 + x2 = - (-2), x1 + x2 = 2

Now it is important to think about what numbers in the product give one? Naturally this 1 * 1 And -1 * (-1) . From these numbers we select those that correspond to the expression x1 + x2 = 2, of course - this is 1 + 1. So we found the roots of the equation: x1 = 1, x2 = 1. This is easy to check if we substitute x^2 into the expression - 2x + 1 = 0.