algebra help, unit 7

Question

Find a value for k so that x+8  divides evenly into x^2+kx+8

Answer 1

Since x+8 divides evenly into x^2+kx+8, x+ 8 is a factor of x^2+kx+8, therefore, -8 is a root of x^2+kx+8 = 0.

Plug -8 into x^2+kx+8 = 0, we have 64 -8k + 8 =0, then k = 9.

Answer 2

factor the numerator so one of the binomial is x+8.

(x+ a)(x+8) = x^2 + x(8+a) + 8a

8a = 8 , a = 1

so linear term (8+a)x = 9x

k = 9

Answer 3

x² + kx + 8 = (x + 8)(x + 1)

x² + kx + 8 = x² + 9x + 8