Question

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

• A. $(c+d+a+b)(c+d-a+b)$
• B. $(c+d+a-2b)(c+2d-a+b)$
• C. $(c+d+a-b)(c+d-a+b)$
• D. $(c+d^2+a-b)(c+d-a+b^2)$

Answer 1

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

$2(ab+cd)-a^2-b^2+c^2+d^2$
$=c^2+2cd+d^2-a^2+2ab-b^2$
$=c^2+2cd+d^2-(a^2-2ab+b^2)$
We know that,
{ $a^2-b^2=(a+b)(a-b)$}
$=(c+d{)}^2-(a-b{)}^2$
$=\left[\right(c+d)+(a-b\left)\right]\left[\right(c+d)-(a-b\left)\right]$
$=(c+d+a-b)(c+d-a+b)$