# Factorization of perfect square trinomial

In this chapter we will learn to factorize polynomial given in the form of perfect square trinomial.

To solve these question you need to memorize some formulas mentioned below.

## Factorizing perfect square trinomial

When the polynomial is expressed in the form of perfect square trinomial, then we can apply following formulas;

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

Please remember this formula as it will be used again and again in math problems.

The trick is to identify the structure of given polynomial and if found suitable, apply the above formula.

Let us see some examples for further understanding.

## Factorizing perfect square trinomial – solved problems

Example 01
\mathtt{\Longrightarrow \ 25x^{2} +10xy\ +4y^{2}}

Solution
The above expression can be written as;

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

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

Where;
a = 5x
b = 2y

Applying the formula we get;

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

Hence, the polynomial is factorized to \mathtt{( 5x\ +\ 2y)^{2}}

Example 02
\mathtt{\Longrightarrow \ 81-108x\ +36x^{2}}

Solution
The above expression can be written as;

\mathtt{\Longrightarrow \ ( 9)^{2} -2( 9)( 6x) \ +( 6x)^{2}}

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

Where;
a = 9
b = 6x

Using the above formula;

\mathtt{\Longrightarrow \ ( 9)^{2} -2( 9)( 6x) \ +( 6x)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 9\ -\ 6x)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 9-6x) .( 9-6x)}

Hence, the given polynomial is factorized to (9 – 6x).(9 – 6x)

Example 03
Factorize the below polynomial
\mathtt{\Longrightarrow \ x^{2} +10x\ +24}

Solution
Here number 24 can be written as;
24 = 25 – 1

Rewriting the above polynomial as;

\mathtt{\Longrightarrow \ x^{2} +10x\ +24}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +10x+25-1}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +2.5.x\ +\ ( 5)^{2} -1}

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

Where;
a = x
b = 5

Using the formula, we get;

\mathtt{\Longrightarrow \ x^{2} +2.5.x\ +\ ( 5)^{2} -1}\\\ \\ \mathtt{\Longrightarrow \ ( x+5)^{2} -1^{2}}

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

Where;
a = x + 5
b = 1

We will get;

\mathtt{\Longrightarrow \ ( x+5)^{2} -1^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x+5+1)( x+5-1)}\\\ \\ \mathtt{\Longrightarrow \ ( x+6)( x+4)}

Hence, the polynomial is reduced to (x+6)(x+4)

Example 04
Factorize the below polynomial
\mathtt{\Longrightarrow \ 16x^{4} +\ 48x^{2} y^{2} \ +36y^{4}}

Solution
The above polynomial can be expressed as;

\mathtt{\Longrightarrow \ 16x^{4} +\ 48x^{2} y^{2} \ +36y^{4}}\\\ \\ \mathtt{\Longrightarrow \ \left( 4x^{2}\right)^{2} +2.\left( 4x^{2}\right)\left( 6y^{2}\right) +\left( 6y^{2}\right)^{2}}

Referring the formula;

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

Where;
a = \mathtt{4x^{2}}
b = \mathtt{6y^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ \left( 4x^{2} +6y^{2}\right)^{2}}\\\ \\ \mathtt{\Longrightarrow \ \left( 4x^{2} +6y^{2}\right) .\left( 4x^{2} +6y^{2}\right)}

Hence, the given polynomial is reduced to \mathtt{\left( 4x^{2} +6y^{2}\right) .\left( 4x^{2} +6y^{2}\right)}

Example 05
Factorize the below polynomial
\mathtt{\Longrightarrow \ 4x^{4} +1}

Solution
To use perfect square trinomial formula, we have to do some manipulations.

\mathtt{\Longrightarrow \ 4x^{4} +1}\\\ \\ \mathtt{\Longrightarrow \ \left( 2x^{2}\right)^{2} +1^{2} +4x^{2} -4x^{2}}

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

Where;
a = \mathtt{2x^{2}}
b = 1

\mathtt{\Longrightarrow \ \left( 2x^{2} +1\right)^{2} -4x^{2}}\\\ \\ \mathtt{\Longrightarrow \ \left( 2x^{2} +1\right)^{2} -( 2x)^{2}}

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

\mathtt{\Longrightarrow \ \left( 2x^{2} +1\right)^{2} -( 2x)^{2}}\\\ \\ \mathtt{\Longrightarrow \ \left( 2x^{2} +1+2x\right) \ \left( 2x^{2} +1-2x\right)}

Hence, the above term is the solution.

Example 06
Solve the below polynomial
\mathtt{\Longrightarrow \ 49x^{2} +\frac{1}{16} +\frac{7x}{2}}

Solution
The above polynomial can be expressed as;

\mathtt{\Longrightarrow \ ( 7x)^{2} +\left(\frac{1}{4}\right)^{2} +2.( 7x) .\left(\frac{1}{4}\right)}

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

Where;
a = 7x
b = 1/4

Applying the formula we get;

\mathtt{\Longrightarrow \ \left( 7x+\frac{1}{4}\right)^{2}}

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

## Practice Questions

Factorize the below polynomials using above identities.

\mathtt{( a) \ x^{2} +14x\ +\ 49}\\\ \\ \mathtt{( b) \ x^{2} -22xy\ +\ 121y^{2} \ }\\\ \\ \mathtt{( c) \ x^{2} -\frac{2}{3} x+\frac{1}{9}}\\\ \\ \mathtt{( d) \ ( x+2y)^{2} \ +\ 2( x+2y)( x-3y) +( x-3y)^{2} \ }\\\ \\ \mathtt{( e) \ 169x^{2} -26x+1}

Solutions

\mathtt{( a) \ ( x+7)^{2}}\\\ \\ \mathtt{( b) \ ( x-11y)^{2}}\\\ \\ \mathtt{( c)\left( x-\frac{1}{3}\right)^{2}}\\\ \\ \mathtt{( d) \ 2( x-y)^{2}}\\\ \\ \mathtt{( e) \ ( 13x-1)^{2}}

Next chapter : Factorizing quadratic expression