Question

How do you find all the zeros of `f(x)=2x^3+x^2-8x-4`?

Answer 1

Zeros of `f(x)` are `{2, -2, -1/2}`

Explanation:

To find zeros of `f(x)=2x^3+x^2-8x-4`, first identify factors `-4`, for which `f(x)=0`. If its zero for `x=a`, `(x-a)` will be a factor of `f(x)`.

Factors of `-4` are `{1,-1,2,-2,4,-4}` and it is apparent that for `x=2`, `f(2)=2(2)^3+2^2-8*2-4=16+4-16-4=0`, hence `(x-2)` is a factor of `2x^3+x^2-8x-4`. Dividing latter by former

`2x^3+x^2-8x-4=(x-2)(2x^2+5x+2)`

Now `(2x^2+5x+2)` can be easily factorized by splitting `5x` as `4x+x` - (In a quadratic polynomial `ax^2+bx+c`, one has to identify factors of `ac`, whose sum is `b`. Hence, here factors of `2xx2=4` are `4`and `1`, whose sum is `5`).

`(2x^2+5x+2)=(2x^2+4x+x+2)=2x(x+2)+1(x+2)=(2x+1)(x+2)`

Hence, factors of `f(x)=2x^3+x^2-8x-4` are `(x-2)(x+2)(2x+1)`.

As zeros of `f(x)` are given by `x-2=0`, `x+2=0` and `2x+1=0`

Hence, zeros of `f(x)` are `{2, -2, -1/2}`