Question

Factorize $(a^2-b^2)(c^2-d^2)-4abcd$

• A. $(ac-bc+ad+bd)(ac-bc-ad-bd)$
• B. $(ac-bc-ad+bd)(ac-bc+ad-bd)$
• C. $(ac-bc+ad+bd)(ac+bc-ad-bd)$
• D. $(ac+bc-ad+bd)(ac+bc-ad-bd)$

Answer 1

The correct option is A

$(ac-bc+ad+bd)(ac-bc-ad-bd)$

Given that $(a^2-b^2)(c^2-d^2)-4abcd$
Multiplying and opening the bracket , we get
$a^2{c}^2-a^2{d}^2-b^2{c}^2+b^2{d}^2-4abcd$
$(a^2{c}^2+b^2{d}^2-2abcd)-(a^2{d}^2+b^2{c}^2+2abcd$)
(Splitting 4abcd as 2abcd + 2abcd and regrouping)
$(ac-bc{)}^2-(ad+bd{)}^2$
( using the identity $a^2-b^2$, we get )
$ac-bc+ad+bd\left)\right(ac-bc-ad-bd)$