Factoring a polynomial is the product of the two or more polynomials. Learn How to Factorize the Difference of Two Squares in this article. Break down all the huge algebraic expressions into small factors with the help of **factorization**. Solved Problems on Factoring the Difference of Two Squares are explained clearly along with the solutions. Visit all factorization problems and get complete knowledge of the factorization concept.

## Solved Problems on How to Factorize the Difference of Two Squares

1. Factorize the following algebraic expressions

(i) m^{2}Â – 121

Solution:

Given expression is m^{2}Â – 121

Rewrite the above expression.

m^{2} – (11)^{2}^{Â }

The above equation m^{2} – (11)^{2}^{Â }is in the form of a^{2}Â â€“ b^{2}.

m^{2} – (11)^{2}^{Â }

Now, apply the formula of a^{2}Â â€“ b^{2} = (a + b) (a â€“ b), where a = m and b = 11

(m + 11) (m – 11)

The final answer is (m + 11) (m – 11)

(ii) 49a^{2} – 16b^{2}

Solution:

Given expression is 49a^{2} – 16b^{2}

Rewrite the above expression.

(7a)^{2} – (4b)^{2}^{Â }

The above equation (7a)^{2} – (4b)^{2}^{Â }^{Â }is in the form of a^{2}Â â€“ b^{2}.

(7a)^{2} – (4b)^{2}^{Â }

Now, apply the formula of a^{2}Â â€“ b^{2} = (a + b) (a â€“ b), where a = 7a and b = 4b

(7a + 4b) (7a – 4b)

The final answer is (7a + 4b) (7a – 4b)

2. Factor the following

(i) 48m^{2} – 243n^{2}

Solution:

Given expression is 48m^{2} – 243n^{2}

Take 3 common

3{16m^{2} – 81n^{2}}

Rewrite the above expression.

3{(4m)^{2} – (9n)^{2}}^{Â }

The above equation {(4m)^{2} – (9n)^{2}}^{Â }^{Â }^{Â }is in the form of a^{2}Â â€“ b^{2}.

{(4m)^{2} – (9n)^{2}}^{Â }

Now, apply the formula of a^{2}Â â€“ b^{2} = (a + b) (a â€“ b), where a = 4m and b = 9n

(4m + 9n) (4m – 9n)

3{(4m + 9n) (4m – 9n)}

The final answer is 3{(4m + 9n) (4m – 9n)}

(ii) 3a^{3} – 48a

Solution:

Given expression is 3a^{3} – 48a

Take 3 common

3a{a^{2} – 16}

Rewrite the above expression.

3a{(a)^{2} – (4)^{2}}^{Â }

The above equation {(a)^{2} – (4)^{2}}^{Â }^{Â }^{Â }^{Â }is in the form of a^{2}Â â€“ b^{2}.

{(a)^{2} – (4)^{2}}^{Â }

Now, apply the formula of a^{2}Â â€“ b^{2} = (a + b) (a â€“ b), where a = a and b = 4

(a + 4) (a – 4)

3a{(a + 4) (a – 4)}

The final answer is 3a{(a + 4) (a – 4)}

3. Factor the expressions

(i) 25(a + 3b)^{2} – 16 (a – 3b)^{2}

Solution:

Given expression is 25(a + 3b)^{2} – 16 (a – 3b)^{2}

Rewrite the above expression.

{[5(a + 3b)]^{2} – [4 (a – 3b)]^{2}}^{Â }

The above equation {[5(a + 3b)]^{2} – [4 (a – 3b)]^{2}}^{Â }is in the form of a^{2}Â â€“ b^{2}.

{[5(a + 3b)]^{2} – [4 (a – 3b)]^{2}}

Now, apply the formula of a^{2}Â â€“ b^{2} = (a + b) (a â€“ b), where a = 5(a + 3b) and b = 4 (a – 3b)

(5(a + 3b) + 4 (a – 3b)) (5(a + 3b) – [4 (a – 3b)])

(5a + 15b + 4a – 12b) (5a + 15b – 4a + 12b)

(9a + 3b) (a + 27b)

3(3a + b) (a + 27b)

The final answer is 3(3a + b) (a + 27b)

(ii) 4x^{2} – 16/(25x^{2})

Solution:

Given expression is 4x^{2} – 16/(25x^{2})

Rewrite the above expression.

{[2x]^{2} – [4/5x]^{2}}

The above equation {[2x]^{2} – [4/5x]^{2}} is in the form of a^{2}Â â€“ b^{2}.

{[2x]^{2} – [4/5x]^{2}}

Now, apply the formula of a^{2}Â â€“ b^{2} = (a + b) (a â€“ b), where a = 2x and b = 4/5x

(2x + 4/5x) (2x – 4/5x)

The final answer is (2x + 4/5x) (2x – 4/5x)