Factoring Quadratic Polynomials – Cases and Examples

Factoring quadratic polynomials consist of decomposing the quadratic equation to form a product of its factors. We can consider factoring as the reverse process of the multiplication distribution. We will look at two cases of factorization of quadratic polynomials: when the leading coefficient is 1 and when the leading coefficient is greater than 1.

ALGEBRA
factoring quadratic polynomials

Relevant for

Learning to factor quadratic polynomials with examples.

See examples

ALGEBRA
factoring quadratic polynomials

Relevant for

Learning to factor quadratic polynomials with examples.

See examples

How to factor quadratic polynomials?

For quadratic equations of the type a{{x}^2}+bx+c=0, we can follow the following steps:

Step 1: Expand the expression and eliminate all fractions if necessary.

Step 2: Move all terms to the left side of the equals sign.

Step 3: Factor the equation by separating the term from the middle.

Step 4: Solve the linear equation for each factor.

EXAMPLE 1

Solve the equation 2({{x}^2+1)=5x.

Solution: We start by expanding and moving all the terms to the left of the equals sign:

2({{x}^2+1)=5x

⇒  2{{x}^2+2=5x

⇒  2{{x}^2-5x-2=0

Then, we separate the middle term to factor:

⇒  2{{x}^2-4x-x-2=0

⇒  (x-2)(2x-1)=0

Now, we can set each factor equal to zero and solve:

⇒  (x-2)=0   y   (2x-1)=0

⇒  x=2   y   x=\frac{1}{2}

EXAMPLE 2

Solve the equation 3{{x}^2-8x-5=0.

Solution: In this case, we have nothing to expand and all the terms are already on the left, so we start by separating the middle terms and factoring:

3{{x}^2-8x-5=0

⇒  3x(x-3)+1(x-3)=0

⇒  (3x+1)(x-3)=0

Now, we can set each factor equal to zero and solve:

⇒  (3x+1)=0   y   (x-3)=0

⇒  x=-\frac{1}{3}   y   x=3

EXAMPLE 3

Solve the equation {{(2x-3)}^2=25.

Solution: We start by expanding the equation to obtain:

4{{x}^2-12x+9-25=0

⇒  4{{x}^2}-12x-16=0

Now, we can divide the entire expression by 4 and factor:

⇒  {{x}^2}-3x-4=0

⇒  (x-4)(x+1)=0

Now, we can set each factor equal to zero and solve:

⇒  (x-4)=0   y   (x+1)=0

⇒  x=4   y   x=-1There are several methods for factoring quadratic polynomials. Our approach here will be to factor quadratic polynomials in which the coefficient of x² is either 1 or greater than 1.

Therefore, we will use trial and error to get the correct factors to factor the quadratic equations.


Factor quadratic polynomials when the coefficient of x² is equal to 1

To factor a quadratic equation of the form {{x}^2}+bx+c, where the leading coefficient is 1, we need to identify two numbers so that their product equals c and their sum equals b.

Case 1: When b and c are both positive

EXAMPLE 1

Factor and solve the equation {{x}^2}+5x+6=0.

Solution: We list the factors of 6: 1×6,  2×3.

Now, we identify the factors with a product of 6 and a sum of 5:

1+6≠5

2+3=5

We check the factors using the distributive property:

(x+2)(x+3)={{x}^2}+2x+3x+6

={{x}^2}+5x+6

Therefore, the factored equation is:

(x+2)(x+3)=0

Now, we can set each factor equal to zero and solve:

⇒  (x+2)=0   y   (x+3)=0

⇒  x=-2   y   x=-3

EXAMPLE 2

Factor and solve the equation {{x}^2}+8x+16=0.

Solution: We identify the factors that produce a product of 16 and a sum of 8:

4×4=16   y   4+4=8

We check the factors using the distributive property:

(x+4)(x+4)={{x}^2}+4x+4x+16

={{x}^2}+8x+16

The factored equation is:

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

We set each factor equal to zero and solve:

⇒  (x+4)=0   y   (x+4)=0

⇒  x=-4

Case 2: When b is positive and c is negative

EXAMPLE 1

Factor and solve the equation {{x}^2}+4x-5=0.

Solution: We start by writing the factors of -5:  -1×5,  1×-5.

We identify the factors that produce a product of -5 and a sum of 4:

-1+5=4

1-5≠4

We check the factors using the distributive property:

(x-1)(x+5)={{x}^2}-x+5x+-5

={{x}^2}+4x-5

The factored equation is:

(x-1)(x+5)=0

We set each factor equal to zero and solve:

⇒  (x-1)=0   y   (x+5)=0

⇒  x=1   y    x=-5

Case 3: When b and c are both negative

EXAMPLE 1

Factor and solve the equation {{x}^2}-2x-8=0.

Solution: We start by writing the factors of -8:  -1×8,  1×-8,  -2×4, 2×-4.

We identify the factors that produce a product of -8 and a sum of -2:

2-4=-2

We check the factors using the distributive property:

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

={{x}^2}-2x-8

The factored equation is:

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

We set each factor equal to zero and solve:

⇒  (x+2)=0   y   (x-4)=0

⇒  x=-2   y    x=4

Case 4: When b is negative and c is positive

EXAMPLE 1

Factor and solve the equation {{x}^2}-7x+10=0.

Solution: We start by writing the factors of 10:  -1×-10,  -2×-5.

We identify the factors that produce a product of 10 and a sum of -7:

-2-5=-7

-1-10≠-7

We check the factors using the distributive property:

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

={{x}^2}-7x+10

The factored equation is:

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

We set each factor equal to zero and solve:

⇒  (x-2)=0   y   (x-5)=0

⇒  x=2   y    x=5


Factor quadratic polynomials when the coefficient of x² is greater than 1

In many cases, the leading coefficient will be different from 1. This means that we will not be able to factor these quadratic equations with the common methods that we have just seen.

Therefore, we must 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

Factor and solve the equation 2{{x}^2}-14x+20=0.

Solution: In this case, we start by looking for whether we have common factors in the expression. Here, we can extract the 2:

⇒   2({{x}^2}-7x+10)=0

Now, we can find the factors of ({{x}^2}-7x+10).Therefore, we write the factors of 10: -1×-10 ,  -2×-5.

We identify the factors that produce a sum of -7:

-2-5=-7

-1-10≠-7

We check the factors using the distributive property:

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

=2({{x}^2}-7x+10)

=2{{x}^2}-14x+20

The factored equation is:

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

We set each factor equal to zero and solve:

⇒  (x-2)=0   y   (x-5)=0

⇒  x=2   y    x=5

EXAMPLE 2

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

Solution: We write the factors of 7 and 11:

1×7=7

1×11=11

We apply the distributive property to verify the factors:

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

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

The factored equation is:

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

We set each factor equal to zero and solve:

⇒  (7x+11)=0   y   (x+1)=0

⇒  x=-\frac{11}{7}   y    x=-1

EXAMPLE 3

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

Solution: Following the same process as the previous examples, we can obtain the factored equation:

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

We set each factor equal to zero and solve:

⇒  (3x+1)=0   y   (3x+1)=0

⇒  x=-\frac{1}{3}

EXAMPLE 4

Factor and solve the equation 6{{x}^2}-7x+2=0.

Solution: We can start by separating the middle term:

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

And now we follow the same process used in the previous examples:

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

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

We set each factor equal to zero and solve:

⇒  (3x-2)=0   y   (2x-1)=0

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


See also

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