Question

What are the zeros of the polynomial function f(x) = x3 − 2x2 − 8x?

Answer 1

$x^3- 2x^2- 8x=0\\ x(x^2-2x-8)=0\\ x=0\\\\ x^2-2x-8=0\\ x^2-2x+1-9=0\\ (x-1)^2=9\\ x-1=3\vee x-1=-3\\ x=4 \vee x-2\\\\ x\in\{-2,0,4\}$

Answer 2

Our strategy will aim to factor the polynomial as much as possible: once completely factored, the polynomial will become a multiplication of polynomials of lower degree:

$f(x) = f_1(x)\cdot f_2(x) \cdot f_3(x)$

and its zeroes will be the ones of its factors.

Since the polynomial has no constant term, you can factor it as follows:

$x^3 - 2x^2 - 8x = x(x^2 - 2x - 8)$

To continue, we must factor the quadratic expression in the parenthesis. A common way to factor expressions like $ax^2+bx+c$ is to find the two solutions $x_1$ and $x_2$ and write the polynomial as $ax^2+bx+c=a(x-x_1)(x-x_2)$.

To find the solutions, we can use the quadratic formula

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

and since in our case $a=1, b=-2,c=-8$, the solving formula becomes

$x_{1,2} = \frac{2\pm\sqrt{4+32}}{2} = \frac{2\pm6}{2} = 1 \pm 3$

So, the two solutions are $x_1 = -2$ and $x_2 = 4$ and we write the polynomial as $(x+2)(x-4)$.

So, the complete factorization is

$x^3 - 2x^2 - 8x = x(x+2)(x-4)$

So, the zeroes of the cubic polynomial we started with are the zeroes of the three polynomials in the factorization: $f_1(x)=0$ yields a solution for $x=0$, $f_2(x)=x+2$ yields a solution for $x=-2$ and $f_3(x)=x-4$ yields a solution for $x=4$.