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 theorem converse of the theorem Vietta. 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.
Expanding polynomials to obtain a product can sometimes seem confusing. But it's not that difficult if you understand the process step by step. The article describes in detail how to factorize quadratic trinomial.
Many people do not understand how to factor a square trinomial and why this is done. At first it may seem like a futile exercise. But in mathematics nothing is done for nothing. The transformation is necessary to simplify the expression and ease of calculation.
A polynomial of the form – ax²+bx+c, called a quadratic trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say it differently: how to decompose quadratic equation.
Interesting! A polynomial is called a square because of its largest degree, the square. And a trinomial - because of the 3 components.
Some other types of polynomials:
- linear binomial (6x+8);
- cubic quadrinomial (x³+4x²-2x+9).
Factoring a quadratic trinomial
First, the expression is equal to zero, then you need to find the values of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. You need to know its formula by heart: D=b²-4ac.
If the result D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated using the formula.
If, when calculating the discriminant, the result is zero, you can use any of the formulas. In practice, the formula is simply shortened: -b / 2a.
Formulas for different meanings discriminants differ.
If D is positive:
If D equal to zero:
Online calculators
On the Internet there is online calculator. It can be used to perform factorization. Some resources provide the opportunity to view the solution step by step. Such services help to better understand the topic, but you need to try to understand it well.
Useful video: Factoring a quadratic trinomial
Examples
We invite you to view simple examples, how to factor a quadratic equation.
Example 1
This clearly shows that the result is two x's because D is positive. They need to be substituted into the formula. If the roots turn out to be negative, the sign in the formula changes to the opposite.
We know the formula for factoring a quadratic trinomial: a(x-x1)(x-x2). We put the values in brackets: (x+3)(x+2/3). There is no number before a term in a power. This means that there is one there, it goes down.
Example 2
This example clearly shows how to solve an equation that has one root.
We substitute the resulting value:
Example 3
Given: 5x²+3x+7
First, let's calculate the discriminant, as in previous cases.
D=9-4*5*7=9-140= -131.
The discriminant is negative, which means there are no roots.
After receiving the result, you should open the brackets and check the result. The original trinomial should appear.
Alternative solution
Some people were never able to make friends with the discriminator. There is another way to factorize a quadratic trinomial. For convenience, the method is shown with an example.
Given: x²+3x-10
We know that we should get 2 brackets: (_)(_). When the expression looks like this: x²+bx+c, at the beginning of each bracket we put x: (x_)(x_). The remaining two numbers are the product that gives "c", i.e. in this case -10. The only way to find out what numbers these are is by selection. The substituted numbers must correspond to the remaining term.
For example, multiplication the following numbers gives -10:
- -1, 10;
- -10, 1;
- -5, 2;
- -2, 5.
- (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
- (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
- (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
- (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.
This means that the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).
Important! You should be careful not to confuse the signs.
Expansion of a complex trinomial
If “a” is greater than one, difficulties begin. But everything is not as difficult as it seems.
To factorize, you first need to see if anything can be factored out.
For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:
3(x²+3x-10). The result is the already well-known trinomial. The answer looks like this: 3(x-2)(x+5)
How to decompose if the term that is in the square is negative? In this case, the number -1 is taken out of brackets. For example: -x²-10x-8. The expression will then look like this:
The scheme differs little from the previous one. There are just a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets that need to be filled in (_)(_). In the 2nd bracket is written x, and in the 1st what is left. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.
The number 3 is given by the numbers:
- -1, -3;
- -3, -1;
- 3, 1;
- 1, 3.
We solve equations by substituting these numbers. The last option is suitable. This means that the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).
Other cases
It is not always possible to convert an expression. With the second method, solving the equation is not required. But the possibility of transforming terms into a product is checked only through the discriminant.
It is worth practicing solving quadratic equations so that when using the formulas there are no difficulties.
Useful video: factoring a trinomial
Conclusion
You can use it in any way. But it’s better to practice both until they become automatic. Also, learning how to solve quadratic equations well and factor polynomials is necessary for those who are planning to connect their lives with mathematics. All the following mathematical topics are built on this.
Let's find the sum and product of the roots of the quadratic equation. Using formulas (59.8) for the roots of the above equation, we obtain
(the first equality is obvious, the second is obtained after a simple calculation, which the reader will carry out independently; it is convenient to use the formula for multiplying the sum of two numbers by their difference).
The following has been proven
Vieta's theorem. The sum of the roots of the given quadratic equation is equal to the second coefficient c opposite sign, and their product is equal to the free term.
In the case of an unreduced quadratic equation, one should substitute the expressions of formula (60.1) into formulas (60.1) and take the form
Example 1. Compose a quadratic equation using its roots:
Solution, a) Find the equation has the form
Example 2. Find the sum of the squares of the roots of the equation without solving the equation itself.
Solution. The sum and product of the roots are known. Let us represent the sum of squared roots in the form
and we get
From Vieta's formulas it is easy to obtain the formula
expressing the rule for factoring a quadratic trinomial.
Indeed, let us write formulas (60.2) in the form
Now we have
which is what we needed to get.
The above derivation of Vieta's formulas is familiar to the reader from a high school algebra course. Another conclusion can be given using Bezout’s theorem and factorization of the polynomial (paragraphs 51, 52).
Let the roots of the equation then be general rule(52.2) the trinomial on the left side of the equation is factorized:
Opening the brackets on the right side of this identical equality, we obtain
and comparing the coefficients at the same powers will give us the Vieta formula (60.1).
The advantage of this derivation is that it can be applied to equations of higher degrees in order to obtain expressions for the coefficients of the equation in terms of its roots (without finding the roots themselves!). For example, if the roots of the given cubic equation
the essence is that according to equality (52.2) we find
(in our case, opening the brackets on the right side of the equality and collecting the coefficients for various degrees we get
Factoring quadratic trinomials refers to 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 and graphic method solutions. 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. Graphical method It is better to use it 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 not integer, but real numbers, which are easy to find if you have a calculator in front of you. Now let's look at 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).
Example 1.1
x 4 + x 3 - 6 x 2.
Solution
We take out x 2
outside the brackets:
.
2 + x - 6 = 0:
.
Roots of the equation:
, .
.
Answer
Example 1.2
Factor the third degree polynomial:
x 3 + 6 x 2 + 9 x.
Solution
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 this we obtain the factorization of the polynomial:
.
Answer
Example 1.3
Factor the fifth degree polynomial:
x 5 - 2 x 4 + 10 x 3.
Solution
We take out x 3
outside the 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:
.
Answer
Examples of factoring polynomials using formulas
Examples with biquadratic polynomials
Example 2.1
Factor the biquadratic polynomial:
x 4 + x 2 - 20.
Solution
Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).
;
.
Answer
Example 2.2
Factor the polynomial that reduces to a biquadratic:
x 8 + x 4 + 1.
Solution
Let's 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 reflexive polynomial
Factor the reciprocal polynomial:
.
Solution
A reciprocal polynomial has odd degree. Therefore it has root x = - 1
. Divide the polynomial by x -(-1) = x + 1
.
.
, ;
;
;
.
Answer
As a result we get:
Let's make a substitution:
Examples of factoring polynomials with integer roots
.
Solution
Example 3.1
6
-6, -3, -2, -1, 1, 2, 3, 6
.
Factor the polynomial:;
Let's assume that the equation;
(-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
x 1 = 1
6 3 - 6 6 2 + 11 6 - 6 = 60 2 = 2
6 3 - 6 6 2 + 11 6 - 6 = 60 3 = 3
.
So, we found three roots:
.
Answer
, x
Examples of factoring polynomials with integer roots
.
Solution
Example 3.1
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 2
Example 3.2
-2, -1, 1, 2
.
has at least one whole root. Then it is a divisor of the number
(member without x). That is, the whole root can be one of the numbers: 6
;
We substitute these values one by one: 0
;
(-2) 4 + 2·(-2) 3 + 3·(-2) 3 + 4·(-2) + 2 =;
(-1) 4 + 2·(-1) 3 + 3·(-1) 3 + 4·(-1) + 2 =
.
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 = 12 2
Example 3.2
1, 2, -1, -2
.
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 = 54 -1
:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2
= -1
Let's substitute x =
.
So, we have found another root x 2 + 2 = 0 .