# Factoring difference of two squares

In this chapter we will learn to factorize polynomial given in the form of difference of two squares.

To solve this problem, you need to remember and apply one simple formula.

## Factorizing polynomial with difference of two squares

When polynomial is in the pattern of difference of two squares, you can use the following formula for factorization.

\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }

Please memorize this formula as you have to use it multiple times to solve algebra problems.

Given below are some solved examples for further clarity.

## Difference of two squares – Solved Problems

Example 01
Factorize the below polynomial
\mathtt{\Longrightarrow \ 144\ -\ y^{2}}

Solution
The above polynomial can be represented as;

\mathtt{\Longrightarrow \ 144\ -\ y^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 12)^{2} -y^{2}}

Referring the formula;
\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }

Where;
a = 12
b = y

Using the formula, we get;

\mathtt{\Longrightarrow \ ( 12)^{2} -y^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 12-y) \ ( 12+y)}

Hence, the above expression is the solution.

Example 02
\mathtt{\Longrightarrow \ x^{4} \ -\ y^{4}}

Solution
\mathtt{\Longrightarrow \ x^{4} \ -\ y^{4}}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2}\right)^{2} -\left( y^{2}\right)^{2}}

Referring the formula;
\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }

Where;
a = \mathtt{x^{2}}
b = \mathtt{y^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ \left( x^{2} -y^{2}\right)\left( x^{2} +y^{2}\right)}

Again using the same formula for \mathtt{x^{2} -y^{2}}

\mathtt{\Longrightarrow \ \left( x^{2} -y^{2}\right)\left( x^{2} +y^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x-y)( x+y)\left( x^{2} +y^{2}\right)}

Hence, the above expression is simplified form.

Example 03
\mathtt{\Longrightarrow \ 25x^{2} -y^{2} +8y-16}

Solution
Simplifying the above expression.

\mathtt{\Longrightarrow \ 25x^{2} -y^{2} +8y-16}\\\ \\ \mathtt{\Longrightarrow \ 25x^{2} -\left( y^{2} -8y+16\right)}\\\ \\ \mathtt{\Longrightarrow \ 25x^{2} -( y-4)^{2}}

The above expression can be rewritten as;

\mathtt{\Longrightarrow \ ( 5x)^{2} -( y-4)^{2}}

Referring the formula;
\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }

Where;
a = 5x
b = y – 4

Using the formula, we get;

\mathtt{\Longrightarrow \ ( 5x)^{2} -( y-4)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 5x+( y-4)) .( 5x-( y-4))}\\\ \\ \mathtt{\Longrightarrow \ ( 5x+y-4) \ ( 5x-y+4)}

Hence, the above expression is final solution.

Example 04
Factorize the below polynomial
\mathtt{\Longrightarrow \ a^{3} -225a}

Solution
\mathtt{\Longrightarrow \ a^{3} -225a}\\\ \\ \mathtt{\Longrightarrow \ a\ \left( a^{2} -225\right)}\\\ \\ \mathtt{\Longrightarrow \ a\ \left( a^{2} -15^{2}\right)}

Using the formula;
\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }

\mathtt{\Longrightarrow \ a\ \left( a^{2} -15^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ a\ ( a-15)( a+15)}

Hence, the above expression is the solution.

Example 05
Factorize the below expression
\mathtt{\Longrightarrow \ ( x-10)^{2} -( x-7)^{2}}

Solution
Referring the formula;
\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }

Where;
a = x – 10
b = x – 7

Putting the values we get;

\mathtt{\Longrightarrow \ ( x-10)^{2} -( x-7)^{2}}\\\ \\ \mathtt{\Longrightarrow \ (( x-10) +( x-7)) .(( x-10) -( x-7))}\\\ \\ \mathtt{\Longrightarrow \ ( x-10+x-7) .\ ( x-10-x+7)}\\\ \\ \mathtt{\Longrightarrow \ ( 2x-17) .\ ( -3)}\\\ \\ \mathtt{\Longrightarrow \ ( -3) \ ( 2x-17)}

Hence, -3.(2x-17) is the factorized form of given polynomial.

Example 06
Factorize the below polynomial
\mathtt{\Longrightarrow \ 196x^{2} -( y-z)^{2}}

Solution
The above expression can written as;
\mathtt{\Longrightarrow \ ( 14x)^{2} -( y-z)^{2}}

Referring the formula;
\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }

Where;
a = 14x
b = (y – z)

Using the formula, we get;

\mathtt{\Longrightarrow \ ( 14x+( y-z)) \ .( 14x-( y-z))}\\\ \\ \mathtt{\Longrightarrow \ ( 14x+y-z) .( 14x-y+z)}

Hence, the above expression is factorized form of given polynomial.

Example 07
Factorize the below polynomial
\mathtt{\Longrightarrow \ x^{8} -y^{8}}

Solution
\mathtt{\Longrightarrow \ \left( x^{4}\right)^{2} -\left( y^{4}\right)^{2}}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{4} +y^{4}\right)\left( x^{4} -y^{4}\right)}\\\ \\ \mathtt{\Longrightarrow \ \ \left( x^{4} +y^{4}\right) \ \left(\left( x^{2}\right)^{2} -\left( y^{2}\right)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ \ \left( x^{4} +y^{4}\right)\left( x^{2} +y^{2}\right)\left( x^{2} -y^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{4} +y^{4}\right)\left( x^{2} +y^{2}\right)\left( x^{2} -y^{2}\right)( x-y)( x+y)}

Hence, the above expression is factorized form.

## Difference of two square polynomials – Problems

Factorize the below polynomials

\mathtt{( i) \ 225x^{2} -121y^{2}}\\\ \\ \mathtt{( ii) \ x^{2} -400\ }\\\ \\ \mathtt{( iii) \ 361-y^{2}}\\\ \\ \mathtt{( iv) \ a^{4} -81}\\\ \\ \mathtt{( v) \ 1-x^{2}}

Solutions

\mathtt{( i) \ ( 15x-11y)( 15x+11y)}\\\ \\ \mathtt{( ii) \ ( x-20)( x+20)}\\\ \\ \mathtt{( iii) \ ( 19-y) \ ( 19+y)}\\\ \\ \mathtt{( iv) \ \left( a^{2} +9\right)( a-3)( a+3)}\\\ \\ \mathtt{( v) \ ( 1-x)( 1+x)}\

Next chapter : Factorizing perfect square trinomial