Factoring Quadratic Equations with Examples

Factoring quadratic equations consists of rewriting the quadratic equation to form a product of its factors. Factoring can be considered as the reverse process of the multiplication distribution.

Here, we will learn about two cases of factoring quadratic equations. The first case is used when the quadratic equations have a leading coefficient of 1 and the second case is used when the quadratic equations have a leading coefficient that is greater than 1

ALGEBRA
factoring quadratic polynomials

Relevant for

Learning to factor quadratic equations with examples.

See examples

ALGEBRA
factoring quadratic polynomials

Relevant for

Learning to factor quadratic equations with examples.

See examples

Factoring quadratic equations when the coefficient of x² is equal to 1

This is the easiest case of factoring quadratic equations. To factor quadratic equations of the form {{x}^2}+bx+c, where the leading coefficient is 1, we have to find two numbers so that when multiplying them, we obtain the constant c and when adding them, we obtain the coefficient b.

First case: When b and c are both positive

EXAMPLE 1

Factor and solve the quadratic equation {{x}^2}+6x+8=0.

Solution: We can form a list with the factors of 8: 1×8,  2×4.

We have to find the factors that have a sum of 6 and a product of 8:

1+8≠6

2+4=6

We can verify these factors using the distributive property:

(x+2)(x+4)={{x}^2}+2x+4x+8

={{x}^2}+6x+8

Therefore, we have already factored into the equation:

(x+2)(x+4)=0

Now, we solve for each factor:

⇒  (x+2)=0   and   (x+4)=0

⇒  x=-2   and   x=-4

EXAMPLE 2

Solve the equation {{x}^2}+8x+15=0 using factoring.

Solution: The factors with a product of 15 and a sum of 8 are:

3×5=15   and   3+5=8

We use the distributive property to verify:

(x+3)(x+5)={{x}^2}+3x+5x+15

={{x}^2}+8x+15

The factored equation is:

(x+3)(x+5)=0

We solve each factor:

⇒  (x+3)=0   and   (x+5)=0

⇒  x=-3   and   x=-5

Second case: When b is positive and c is negative

EXAMPLE

Solve the equation {{x}^2}+6x-7=0 by factoring.

Solution: The factors of -7 are:  -1×7,  1×-7.

We find the factors with a product of -7 and a sum of 6:

-1+7=6

1-7≠6

We verify with the distributive property:

(x-1)(x+7)={{x}^2}-x+7x-7

={{x}^2}+6x-7

We have already obtained the factorization of the equation:

(x-1)(x+7)=0

We solve for each factor:

⇒  (x-1)=0   and   (x+7)=0

⇒  x=1   and    x=-7

Third case: When b and c are both negative

EXAMPLE 1

Factor and solve the equation {{x}^2}-3x-10=0.

Solution: The factors of -10 are:  -1×10,  1×-10,  -2×5, 2×-5.

We identify the factors that produce a product of -10 and a sum of -3:

2-5=-3

Using the distributive property, we can verify the factors:

(x+2)(x-5)={{x}^2}+2x-5x-10

={{x}^2}-3x-10

Therefore, the factorization of the equation is:

(x+2)(x-5)=0

We solve for each factor:

⇒  (x+2)=0   and   (x-5)=0

⇒  x=-2   and    x=5

Fourth case: When b is negative and c is positive

EXAMPLE

Solve the equation {{x}^2}-9x+14=0 by factoring.

Solution: We start by finding the factors of 14:  -1×-14,  -2×-7.

The factors with a product of 14 and a sum of -9 are:

-2-7=-9

-1-14≠-9

Checking with the distributive property, we have:

(x-2)(x-7)={{x}^2}-2x-7x+14

={{x}^2}-9x+14

The factorization is:

(x-2)(x-7)=0

Solving for each factor, we have:

⇒  (x-2)=0   and   (x-7)=0

⇒  x=2   and    x=7

Factoring quadratic equations when the coefficient of x² is greater than 1

The methods we have just seen do not work when the leading coefficient is different from 1. For these cases, we have to take into account the coefficient of {{x}^2} and the factors of c to find numbers that have a sum equal to b.

EXAMPLE 1

Solve the equation 2{{x}^2}-16x+30=0 by factoring.

Solution: We have to determine if there are common factors in the expression. In this case, we can extract the 2:

⇒   2({{x}^2}-8x+15)=0

We can factor ({{x}^2}-8x+15). The factors of 15 are: -1×-15 ,  -3×-5.

We determine which factors produce a sum of -8:

-3-5=-8

-1-15≠-8

We use the distributive property to verify:

2(x-3)(x-5)=2({{x}^2}-3x-5x+15)

=2({{x}^2}-8x+15)

=2{{x}^2}-16x+30

The factored equation is:

2(x-3)(x-5)=0

We solve for each factor:

⇒  (x-3)=0   and   (x-5)=0

⇒  x=3   and   x=5

EXAMPLE 2

Factor and solve the quadratic equation 7{{x}^2}+18x+11=0.

Solution: We have to find the factors of 7 and 11:

1×7=7

1×11=11

Now, we use the distributive property to check the factors:

(7x+1)(x+11)\ne 7{{x}^2}+18x+11

(7x+11)(x+1)=7{{x}^2}+18x+11

The factorization of the quadratic equation is:

(7x+11)(x+1)=0

We solve for each factor:

⇒  (7x+11)=0   and   (x+1)=0

⇒  x=-\frac{11}{7}   and    x=-1

EXAMPLE 3

Factor and solve the equation 9{{x}^2}+6x+1=0.

Solution: Using the same process as in the previous exercises, we can obtain the factorization of the quadratic equation:

(3x+1)(3x+1)=0

We solve for each factor:

⇒  (3x+1)=0   and   (3x+1)=0

⇒  x=-\frac{1}{3}

EXAMPLE 4

Solve the equation 6{{x}^2}-7x+2=0 by factoring.

Solution: In this case, it is convenient that we start by separating the middle term:

6{{x}^2}-4x-3x+2=0

Following the same process as in the previous exercises, we have:

2x(3x-2)-1(3x-2)=0

(3x-2)(2x-1)=0

Solving for each factor, we have:

⇒  (3x-2)=0   and   (2x-1)=0

⇒  x=\frac{2}{3}    and    x=\frac{1}{2}


See also

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