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
The factorization of square trinomials is carried out by solving 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 of any in a known way. 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. Examples illustrate the factorization of a square 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.
Similarly, we 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 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).
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.
Square trinomial is called a trinomial of the form a*x 2 +b*x+c, where a,b,c are some arbitrary real numbers, and x is a variable. Moreover, the number a should not be equal to zero.
The numbers a,b,c are called coefficients. The number a is called the leading coefficient, the number b is the coefficient of x, and the number c is called the free term.
Root of a square trinomial a*x 2 +b*x+c is any value of the variable x such that the square trinomial a*x 2 +b*x+c vanishes.
In order to find the roots of a quadratic trinomial, it is necessary to solve a quadratic equation of the form a*x 2 +b*x+c=0.
How to find the roots of a quadratic trinomial
To solve this, you can use one of the known methods.
- 1 way.
Finding the roots of a square trinomial using the formula.
1. Find the value of the discriminant using the formula D =b 2 -4*a*c.
2. Depending on the value of the discriminant, calculate the roots using the formulas:
If D > 0, then the square trinomial has two roots.
x = -b±√D / 2*a
If D< 0, then the square trinomial has one root.
If the discriminant is negative, then the quadratic trinomial has no roots.
- Method 2.
Finding the roots of a quadratic trinomial by isolating the perfect square. Let's look at the example of the given quadratic trinomial. A reduced quadratic equation whose equation is by the leading coefficient equal to one.
Let's find the roots of the quadratic trinomial x 2 +2*x-3. To do this, we solve the following quadratic equation: x 2 +2*x-3=0;
Let's transform this equation:
On the left side of the equation there is a polynomial x 2 +2*x, in order to represent it as a square of the sum we need there to be another coefficient equal to 1. Add and subtract 1 from this expression, we get:
(x 2 +2*x+1) -1=3
What can be represented in parentheses as the square of a binomial
This equation breaks down into two cases: either x+1=2 or x+1=-2.
In the first case, we get the answer x=1, and in the second, x=-3.
Answer: x=1, x=-3.
As a result of the transformations, we need to get the square of the binomial on the left side, and a certain number on the right side. The right side should not contain a variable.
Example 1.1
x 4 + x 3 - 6 x 2.
Solution
We take out x 2
outside of 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 here 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 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:
.
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 one:
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 recurrent 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. As a result we get:
.
Let's make a substitution:
, ;
;
;
.
Answer
Examples of factoring polynomials with integer roots
Example 3.1
Factor the polynomial:
.
Solution
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
.
Answer
Example 3.2
Factor the polynomial:
.
Solution
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
.
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:
.
Since the equation x 2 + 2 = 0 has no real roots, then the factorization of the polynomial has the form.
Square trinomial ax 2 +bx+c can be factorized into linear factors using the formula:
ax 2 +bx+c=a (x-x 1)(x-x 2), Where x 1, x 2- roots of quadratic equation ax 2 +bx+c=0.
Factor the quadratic trinomial into linear factors:
Example 1). 2x 2 -7x-15.
Solution. 2x 2 -7x-15=0.
a=2; b=-7; c=-15. This general case for a complete quadratic equation. Finding the discriminant D.
D=b 2 -4ac=(-7) 2 -4∙2∙(-15)=49+120=169=13 2 >0; 2 real roots.
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
2x 2 -7x-15=2 (x+1.5)(x-5)=(2x+3)(x-5). We introduced this trinomial 2x 2 -7x-15 2x+3 And x-5.
Answer: 2x 2 -7x-15= (2x+3)(x-5).
Example 2). 3x 2 +2x-8.
Solution. Let's find the roots of the quadratic equation:
a=3; b=2;c=-8. This special case for a complete quadratic equation with an even second coefficient ( b=2). Finding the discriminant D 1.
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
We introduced the trinomial 3x 2 +2x-8 as a product of binomials x+2 And 3x-4.
Answer: 3x 2 +2x-8 =(x+2)(3x-4).
Example 3). 5x 2 -3x-2.
Solution. Let's find the roots of the quadratic equation:
a=5; b=-3; c=-2. This is a special case for a complete quadratic equation with the following condition: a+b+c=0(5-3-2=0). In such cases first root is always equal to one, and second root equal to the quotient of the free term divided by the first coefficient:
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
5x 2 -3x-2=5 (x-1)(x+0.4)=(x-1)(5x+2). We introduced the trinomial 5x 2 -3x-2 as a product of binomials x-1 And 5x+2.
Answer: 5x 2 -3x-2= (x-1)(5x+2).
Example 4). 6x 2 +x-5.
Solution. Let's find the roots of the quadratic equation:
a=6; b=1; c=-5. This is a special case for a complete quadratic equation with the following condition: a-b+c=0(6-1-5=0). In such cases first root is always equal to minus one, and second root is equal to the minus quotient of dividing the free term by the first coefficient:
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
We introduced the trinomial 6x 2 +x-5 as a product of binomials x+1 And 6x-5.
Answer: 6x 2 +x-5= (x+1)(6x-5).
Example 5). x 2 -13x+12.
Solution. Let's find the roots of the given quadratic equation:
x 2 -13x+12=0. Let's check if it can be applied. For this let's find the discriminant and make sure that it is a perfect square of an integer.
a=1; b=-13; c=12. Finding the discriminant D.
D=b 2 -4ac=13 2 -4∙1∙12=169-48=121=11 2 .
Let's apply Vieta's theorem: the sum of the roots must be equal to the second coefficient taken with the opposite sign, and the product of the roots must be equal to the free term:
x 1 + x 2 =13; x 1 ∙x 2 =12. It is obvious that x 1 =1; x 2 =12.
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
x 2 -13x+12=(x-1)(x-12).
Answer: x 2 -13x+12= (x-1)(x-12).
Example 6). x 2 -4x-6.
Solution. Let's find the roots of the given quadratic equation:
a=1; b=-4; c=-6. The second coefficient is an even number. Find the discriminant D 1.
The discriminant is not a perfect square of an integer, therefore, Vieta’s theorem will not help us, and we will find the roots using the formulas for the even second coefficient:
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2) and write down the answer.