Question

$63a^2-112b^2$. How do we factorize this using a suitable identity?

• A. $7(3a–4b)(3a+4b)$
• B. $7(3a+b)(a–4b)$
• C. $(3a–4b)(3a–4b)$
• D. $(3a–4b)(3a+4b)$

Answer 1

The correct option is A

$7(3a–4b)(3a+4b)$

Explanation of correct option-

We have-

$63a^2-112b^2=7(9a^2–16b^2)$

$=7[{\left(3a\right)}^2–{\left(4b\right)}^2]$ Using a² – b² = (a – b)(a + b)

$=7(3a–4b)(3a+4b)$

Hence, option a is correct