Question

What are the roots of the polynomial equation x^2+8x-48=0

Answer 1

 Given polynomial x^2+8x-48 = 0

x^2+12x-4x-48 = 0

x(x+12)-4(x+12) = 0

(x+12)(x-4) = 0

x+12 = 0

Subtract 12 from each side.

x+12-12 = 0-12

x = -12

and x-4 = 0

Add 4 to each side.

x-4+4 = 0+4

x = 4

Roots are -12,4.