Question

How do you find a function with a degree of 4 that has zeros at x= -1, 0, 1?

I'm not given a point that the function passes through. :(

Answer 1

f(x)= `x^2`(`x^2`-1)

Explanation:

If x has a zero at -1, then (x+1)=0. Here's one factor of our function.

If x has a zero at 1, then (x-1)=0. Here's another factor of our function.

If x has a zero at 0, then x=0. But this cannot be all, these three factors would only give us a function of the third degree.

So, we can say that `x^2=0`.

Our final function is:

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

Note that this satisfies our four given requirements. If we were to multiply everything through, we'd have a function of the fourth degree (the highest exponent in our function would be 4). f(0)=0, f(-1)=0, and f(1)=0.