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

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\}`