Question

A rectangle's area is $x^2-2x-35$. What are the possible dimensions of the rectangle. (Use method of factorisation)

• A. $(x-7)\text{and}(x-5)$
• B. $(x+7)\text{and}(x-5)$
• C. $(x+7)\text{and}(x+5)$
• D. $(x-7)\text{and}(x+5)$

Answer 1

The correct option is D $(x-7)\text{and}(x+5)$

Given, area of rectangle = $x^2-2x-35$
We know that area of rectangle = length x breadth.
So, if we factorise the given expression and write it as a product of factors, those factors will be the possible dimensions of the rectangle.
Let's factorise $x^2-2x-35$.
Observe that -7x5 = -35 and -7 + 5 = -2
We have,
$x^2-2x-35$ = $x^2-7x+5x-(7\times 5)$
= $x(x-7)+5(x-7)$
= $(x-7)(x+5)$

$\therefore$ area of rectangle = length x breadth
= $(x-7)(x+5)$

Therefore possible dimensions of rectangle are $(x-7)\text{and}(x+5)$.