Question

How do you find all the zeros of `f(x)= x^4 - 4x^3 - 20x^2 + 48x`?

Answer 1

Factors of `f(x)=x^4−4x^3−20x^2+48x`) are `x(x-2)(x+4)(x-6)` and zeros of `f(x)` are `{0,2, -4, 6}`

Explanation:

As there is no independent in `f(x)=x^4−4x^3−20x^2+48x` and hence`x` is factor of `f(x)` and hence `)` is first zero among the zeros of `f(x)`. Factorizing `f(x)`

`f(x)=x(x^3-4x^2-20x+48)`

Now identify factors of `48` i.e. `{1,-1,2,-2,3,-3,4,-4,6,-6,...}`, who make `(x^3-4x^2-20x+48)=0`. As `x=2`, makes `f(x)=0`, and hence `(x-2)` is a factor of `f(x)` and `2` is another zero of `f(x)`.

Factorizing `(x^3-4x^2-20x+48)=(x-2)(x^2-2x-24)`

Now `(x^2-2x-24)`, can be easily factorized as `-24` can be exactly factorized into `-6` and `4`, and hence factorization can proceed as under.

`x^2-2x-24=x^2-6x+4x-24=x(x-6)+4(x-6)=(x+4)(x-6)`

Hence complete factors of `f(x)=x^4−4x^3−20x^2+48x`) are `x(x-2)(x+4)(x-6)` and zeros of `f(x)` are `{0,2, -4, 6}`