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 C

$(c+d+a+b)(c+d-a-b$)

Given
$2(ab+cd)-a^2-b^2+c^2+d^2$
Opening the bracket , we get
$2ab+2cd-a^2-b^2+c^2+d^2$
Re-grouping
$(c^2+d^2+2cd)-(a^2-2ab+b^2)$
Using $(a+b{)}^2=a^2+2ab+b^2$ and $(a-b{)}^2=a^2-2ab+b^2$
$(c+d{)}^2-(a+b{)}^2$
Using difference of squares , we have
$(c+d+a+b)(c+d-a-b$)