Question

How do you find the polynomial function with roots `-1/2, 1/4, 1, 2`?

Answer 1

`8x^4-22x^3+9x^2+7x-2`

Explanation:

Consider the equation `(x-2)(2x+3)=0`.

To figure out what `x` could be, we set each term being multiplied equal to `0`.

`{(x-2=0rarrx=2),(2x+3=0rarrx=-3/2):}`

To answer your question, we use this process in reverse.

If we know one of the roots of the polynomial is `-1/2`, we know `x=-1/2`. Therefore, `x+1/2=0`, but having it's easier if we don't have fractions. We can multiply everything by `2` to see that `2x+1=0`. From this, we know that one of the multiplied terms in our polynomial will be `(2x+1)`.

We can use the same logic for each of the roots:

`{(-1/2rarr(2x+1)),(1/4rarr(4x-1)),(1rarr(x-1)),(2rarr(x-2)):}`

Then, we multiply them:

`f(x)=(2x+1)(4x-1)(x-1)(x-2)`

If we foil, we find that `f(x)=8x^4-22x^3+9x^2+7x-2`.