What is the value of k such that (x^3−x^2+kx+40)÷(x + 5) has a remainder of zero? Then, factor the polynomial completely.
+37146 Long division
x^2 -6x +8
x+5 | x^3 -x^2 + kx + 40
x^3 + 5x^2
- 6x^2 + kx + 40
-6x^2 -30x
(k+30)x + 40 k+ 30 must = 8 for this to have no remainder so k = -22
8x +40
(x+5)(x^2 -6x+8)
(x+5)(x-4)(x-2) is factored
