Question

Factorise $2(ab+cd)-a^2-b^2+c^2+d^2$

• A. $(c+d-a+b)(c+d-a-b)$
• B. $(c+d+a+b)(c+d-a-b)$
• C. $(c-d+a+b)(c-d-a-b)$
• D. $(c+d+a-b)(c-d-a-b)$

Answer 1

The correct option is B $(c+d+a+b)(c+d-a-b)$

$2(ab+cd)-a^2-b^2+c^2+d^2$
= $2ab+2cd-a^2-b^2+c^2+d^2$
Re-arranging the terms , we get
= $c^2+2cd+d^2-(a^2+2ab+b^2)$
We know $(a+b{)}^2=a^2+2ab+b^2$
= $(c+d{)}^2-(a+b{)}^2$
This is of the form $x^2-y^{2}wherex=c+dandy=a+b$
We know $x^2-y^2=(x+y)(x-y)$
= $(c+d+a+b)(c+d-a-b)$