\(\text{Let } f(x) = x^3 + 2x^2 + kx - 6\)
As (x + 1) is a factor, it follows that
\(f(-1) = 0\)
\(\implies (-1)^3 + 2(-1)^2 + k(-1) - 6 = 0\)
\(-1 + 2 -k - 6 = 0\)
\(-k - 5 = 0\)
\(k = -5\)
\(\implies f(x) = x^3 + 2x^2 - 5x - 6\)
Now we can use long division to find what is left of f(x) after it is divided by (x + 1). (Apologies, this is the best way I can represent long division on Brainly at this current time - I hope it's clear)
x^2 + x - 6
x + 1 ( x^3 + 2x^2 - 5x - 6
x^3 + x^2
x^2 - 5x
x^2 + x
-6x - 6
-6x - 6
0
So the remainder when f(x) is divided by (x + 1) is
\(x^2 + x - 6\)
Factorising this we get
\((x + 3)(x - 2)\)
So the three factors of f(x) are (x + 1), (x + 3) and (x - 2).
