Before you start practicing the Difference of Two Squares Concept know the step-by-step process to solve the Factoring Difference of Two Squares. Therefore, students can easily learn all types of factoring problems here. Without any late, begin your practice and finish solving every problem included. Refer to Solved Examples on Difference of Two Squares with Answers Provided.

## Difference of Two Squares Questions

1. m^{4} – (n + r)^{4}

Solution:

Given expression is m^{4} – (n + r)^{4
}Rewrite the given expression in the form of a^{2} – b^{2}.

(m^{2})^{2} – ( (n + r)^{2})^{2
}Now, apply the formula of a^{2} – b^{2} = (a + b) (a – b), where a = m^{2} and b = (n + r)^{2
}[m^{2} + (n + r)^{2}] [m^{2} – (n + r)^{2}]

[m^{2} + n^{2} + r^{2} + 2nr] [(m)^{2} – (n + r)^{2}]

From the above equation, [(m)^{2} – (n + r)^{2}] is in the form of a^{2} – b^{2}.

[(m)^{2} – (n + r)^{2}]

Now, apply the formula of a^{2} – b^{2} = (a + b) (a – b), where a = m and b = (n + r)

[m + (n + r)] [m – (n + r)]^{
}Now, [m^{2} + n^{2} + r^{2} + 2nr] [(m)^{2} – (n + r)^{2}]

[m^{2} + n^{2} + r^{2} + 2nr] [m + (n + r)] [m – (n + r)]

[m^{2} + n^{2} + r^{2} + 2nr] [m + n + r] [m – n – r]

The final answer is [m^{2} + n^{2} + r^{2} + 2nr] [m + n + r] [m – n – r]

2. 4a^{2} – b^{2} + 6b – 9.

Solution:

Given expression is 4a^{2} – b^{2} + 6b – 9.^{
}Rewrite the given expression.

4a^{2} – (b^{2} – 6b + 9)

b^{2} – 6b + 9 is in the form of a^{2} – b^{2} + 2ab where a = b, b = 3

We know that a^{2} – b^{2} + 2ab = (a – b)^{2}

Therefore, b^{2} – 6b + 9 = (b – 3)^{2}

So, 4a^{2} – (b – 3)^{2}

The above equation 4a^{2} – (b – 3)^{2 }is in the form of a^{2} – b^{2}.

[(2a)^{2} – (b – 3)^{2}]

Now, apply the formula of a^{2} – b^{2} = (a + b) (a – b), where a = 2a and b = (b – 3)

(2a + b – 3) {2a – (b – 3)},

(2a + b – 3) (2a – b – 3)

The final answer is (2a + b – 3) (2a – b – 3)

3. 25x^{2} – (4m^{2} – 12mn + 9n^{2})

Solution:

Given expression is 25x^{2} – (4m^{2} – 12mn + 9n^{2})

(4m^{2} – 12mn + 9n^{2}) is in the form of a^{2} – b^{2} + 2ab where a = 2m, b = 3n

We know that a^{2} – b^{2} + 2ab = (a – b)^{2}

Therefore, (4m^{2} – 12mn + 9n^{2}) = (2m – 3n)^{2}

So, 25x^{2} – (2m – 3n)^{2
}The above equation 25x^{2} – (2m – 3n)^{2 }is in the form of a^{2} – b^{2}.

[(5x)^{2} – (2m – 3n)^{2}]

Now, apply the formula of a^{2} – b^{2} = (a + b) (a – b), where a = 5x and b = (2m – 3n)

[5x + (2m – 3n)] [5x – (2m – 3n)]

(5x + 2m – 3n) (5x – 2m + 3n)

The final answer is (5x + 2m – 3n) (5x – 2m + 3n)