# Solving Quadratic Equations by Factoring

By Kathleen Knowles, 07 Oct 2020

The term "quadratic" traces from the Latin word "quad," which means "square."

This is because the variable gets squared(X²).

A quadratic equation is, thus, sometimes referred to as Equation of Degree 2 since the greatest power is 2 (having one or more variables raised to the second power).

The standard form of any quadratic equation must be expressed as AX²+ BX + C≠0, where A, B, and C are values, except that A can't be equal to zero, and X is unknown (yet to be solved).

## Here's All You Need to Know About Solving Quadratic Equations by Factoring

There are, basically, three methods of solving Quadratic Equations by Factoring:

- The Sum product pattern method.
- Grouping method.
- Special product method.

### Use the Sum-Product Method in Solving Quadratic Equations by Factorizing

This method is mainly used by students who find it challenging to use the guessing method, (or the trial and error method). Unlike the trial and error method, the Product Sum Method is generally easier to apply since it identifies an equation that cannot be factored.

This method takes various forms, i.e:

Case 1: X² + BX + C= 0 ( A= 1).

Case 2: AX² + BX + C = 0 ( A ≠1).

#### Case 1

Simply follow these steps when solving an equation using the product sum method:

Step One: Find two integers whose product is C.

Step Two: Give the integers any characters of your choice, for example, M and N.

Step Three: Make one factor ( X + M ) and the other ( X + N).

#### Illustration 1

Find the value of x by factorization.

X² + 16X + 55= 0

Solution:

Find two integers whose product is 15. The table below shows the numbers.

Below are the pairs of the numbers

1 and 5

55 and 11

-1 and -5

-55 and -11

You can then select the pair that has the sum of 16 and product 55.

That pair is 5 and 11.

Therefore, the factors are X² + 16X +55 = (X +5)(X +11) = 0,

Thus;

X= - 5 or X = - 11.

#### Illustration 2

Find the value of X in X² -16X +60 =0

Solution

Identify a duo of integers whose product is 60; the pairs are listed below.

You can have a table with different values for different pairs.

You should then select a pair that has sum -16.

Thus, the pair is:

-6 and – 10.

Therefore;

X² -16X + 60 =(X- 6)(X-10) = 0

Hence X = 6 or X = 10.

#### Case 2

If the equation AX²+BX + C =0 and A≠1 you only need a little extra effort to find the factors using the product sum method.

Here are the steps to follow:

- Identify two integers whose product is AC and sum is B.
- You can name the integers M and N.
- Rewrite the function as a four term expression as below AX² + MX + NX + C.
- Use grouping by pair to factor out the Greatest Common Factor (GCF) in the two terms to get a common parenthesis.

To illustrate this case, let's consider the following examples.

#### Example 1

Find the value of X given that 2X²+ X -10=0

#### Solution

Find two integers whose product AC= (2)×(-10)=-20. You should then draw a table on your working paper to come up with several pairs.

You can now select the pair that has the sum of B = 1. This pair is - 4 and 5.

Rewrite the expression as 2X²- 4X + 5x - 10 = 0

Taking out GCF, we get 2X(X-2)5(x-2).

Now we have the common parenthesis, which is X-2.

Therefore,(2X +5)(X-2)= 0 where X = 2 or X = -5/2

#### Example 2

Calculate the value of X given that 3X²+X-2 =0

#### Solution

Find two integers whose product is AC= 3 ×-2=-6

Next, list the pairs in a tabular form.

Select the pair that has the sum of B=1. This pair is 3 and -2

You should then rewrite the function as 3X²+3X-2X-2= 0.

Taking out GCF, we get:

3X( X+1)-2 ( X+1)=0

Now the common parenthesis is X+1;

Therefore ( 3X-2)(X+1)=0

Thus X = 1 or X =2/3

### How to Solve a Quadratic Equation by Factoring Using Grouping Method

This method involves arranging the terms into smaller groupings with common factors. Use the factoring by grouping method if you can't find the common factor for all the terms.

Further, by taking two terms at the same time, you can get something to divide the terms.

Use this easy procedure in solving the equation by factorizing using the grouping method.

Suppose you are given a general equation AX² +BX + C;

- Find the product of AC.
- Think of two numbers, say Q and P such that QP= AC and Q+P= B
- Rewrite the expression as AX² + QX +PX +C
- Group the expression into two pairs that have a common factor and simplify like this:

First Pair 🙁 AX²+QX)+(PX+C)

Second Pair: X(AX+Q)+(PX+C)

Depending on your selection of P and Q, you will factor out a constant on the second parenthesis, remaining with two identical expressions as shown in the example below:

#### Example 1

Find the value of X given 5X² + 11X +2= 0

#### Solution

First find the product AC;

5 ×2=10

Then think of two factors of 10 that can add up to 11

Next, write 11X in the product of 10 and 1.

Hence 5X²+1X + 10X +2.

You should now group the pairs into two.

(5X²+ 1X) + (10 X +2)

After grouping, take out the common factor.

X (5X+1)+2(5X +1)

Thus, (X +2)(5X+1)=0

Therefore X =-2 or X= -1/5

#### Example 2

Compute the value of X given that X²+2X-24=0

#### Solution

First, find the product AC = 1×-24=-24

Think of two factors, such that their product is -24 and their sum is 2.

Let the factors be -4 and +6

Next, write +2X in the form -4X and 6X

Therefore, the expression becomes X²- 4X +6X-24=0

Pair the equation into 2 terms, thus:

(X² -4X)+(6X-24)

Next, take out the common factor.

X(X-4)+6(X-4)

Now, X-4 becomes the common parenthesis.

Therefore (X+ 6)(X-4)=0

Thus, X=4 or X= -6

### Solving Quadratic Equations by Factorizing Using the Special Product Method

The Special Product Method requires special cases that can be factored quicker. You can do this using two special quadratics:

- Perfect Square Expression
- Difference of Two Squares Expression

### Case 1: Perfect Square Expression

- Check whether the first and the last term are perfect squares
- Check whether the middle term is 2 times the product of the roots of the other terms.

#### Illustration

Identify the value of X given that X²+ 14X +49=0

#### Solution

The first term is X² and the last term is 49; both X² and 49 are perfect squares whose roots are X and 7 respectively.

The middle term, 14X is two times the roots of the other terms.

Thus,

X² +14X+49=(X+7)² = 0

Therefore (X+7) (X+7) =0

This X =-7

### Case 2: Difference of Two Squares Expression

Special Product Method is used here since:

- There are no common factors
- The typical middle term is missing
- The terms present are perfect squares and being subtracted.

#### Example 1

Identify the value of X given that X²-16=0

#### Solution

Note that in the quadratic equation above:

- The middle term is missing.
- The terms present² and 16 are perfect squares and are being subtracted.

Thus:

X²-16=(X-4)(X+4)=0

Therefore X= 4 or X=-4

#### Example 2

Find the value of X given that X2-64=0

The terms X² and 64 are perfect squares and they are subtracted.

Thus X2-64= (X-8)(X+8)

Therefore X = 8 or X-8

### Verdict

Quadratics are considered to be among the most challenging concepts in Mathematics.

Regardless, getting the correct methods and learning how to apply the concepts can make teaching and learning Mathematics fun!

Consequently, knowledge of quadratics is important in everyday life.

For instance, quadratics is used in the determination of profits or even in formulating the speed and velocity of an object.

Further, quadratics have been applied to athletic endeavours like shotput and javelin. Take time to learn the various methods to solve quadratic equations using the Factorizing method and you'll come to love it!

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