Finding Dimensions of a Rectangle When Area is Given

Question

A rectangle has an area of x^2-3x-54. Find area.

Answer 1

A rectangle has an area of x^2-3x-54. Find the perimeter.

If a rectangle has area length*height, then the perimeter is 2*(length+height). So, what is length and height in x^2-3x-54. We need to write this in terms of (x+a)*(x+b).

(x+a)*(x+b) = x^2 + (a+b)*x + a*b

a*b = -54

a+b = -3

We need two numbers, whose multiplication is -54 and summation is -3. Let's factor -54

54 = 1*54

= 2*27

= 3*18

= 6*9

Since a*b is actually -54, then either a or b must be a negative number. So, we need a pair, where one is a negative number AND the sum is -3. Let's try 6 and 9, since the difference between these two numbers is 3.

-6+9 = 3 Nope

+6-9 = -3 Yes!

So, a = 6, b = -9

(x+6)*(x-9) = x^2-3x-54

So, the perimeter is 2*((x+6)+(x-9)) = 2*(2x-3) = 4x-6.