Question

What are the real zeros of f(x) = x3 – 3x2 – 4x? A. 0, 1, 4 B. 1, -1, 4 C. -1, 0, 4 D. 1, 0, -4

Answer 1

Step-by-step explanation:

$f(x) = x^3 - 3x^2 - 4x$

factor it:  (start by  factoring out the x that is common to all 3 )

       $= x\left(x^2-3x-4\right)$

next keep factoring the inside...

       $= x\left(x^2+x-4x-4\right)\\=x \left[x\left(x+1\right)-4\left(x+1\right)\right]\\=\bold{x\left(x+1\right)\left(x-4\right)}$

So, the only way the function $f(x)$ can be $0$ is if:

• either $x=0$, or
• $\left(x+1\right)=0$, or

• $\left(x-4\right)=0$

Answer:

               C. $\{-1,0,4\}$