**Find a polynomial of the specified degree that satisfies the given conditions.**

**Degree 4; zeros −1, 0, 5, 1/5**

**; coefficient of x3 is 21**

**Answer:**

you would start off by setting it up as f(x)=(x+1)(x)(x-5)(5x-1)

That is because when you plug the given roots in the equation this will

make the equation equal to 0 in at least one case and anything

multiplied by 0 is 0. Lets factor it now. (x+1)(x) -> x^2 + x

(x^2 + x)(x-5) -> x^3 + x^2 -5x^2 -5x -> x^3 - 4x^2 - 5x

(x^3 - 4x^2 - 5x)(5x-1) -> 5x^4 - 20x^3 - 25x^2 - x^3 + 4x^2 + 5x

-> 5x^4 - 21x^3 - 21x^2 + 5x

now that is right, but it also needs the coefficient on x^3 to be +21 not -21,

thus we multiply by -1 to fix that

(-1)(5x^4 - 21x^3 - 21x^2 + 5x) -> -5x^4 + 21x^3 + 21x^2 - 5x

There we go and we can do that because 0*(-1) = 0

meaning it didnt change the roots.