Question

How do you write a polynomial function of least degree with integral coefficients that has the given zeros -3, -1/3, 5?

Answer 1

`f(x)=3x^3-5x^2-47x-15`

Explanation:

If the zero is c, the factor is (x-c).

So for zeros of `-3,-1/3, 5`, the factors are

`(x+3)(x+1/3)(x-5)`

Let's take a look at the factor `(x+color(blue)1/color(red)3)`. Using the factor in this form will not result in integer coefficients because `1/3` is not an integer.

Move the `color(red)3` in front of the x and leave the `color(blue)1` in place: `(color(red)3x+color(blue)1)`.

When set equal to zero and solved, both
`(x+1/3)=0` and `(3x+1)=0` result in `x=-1/3`.

`f(x)=(x+3)(3x+1)(x-5)`

Multiply the first two factors.

`f(x)=(3x^2+10x+3)(x-5)`

Multiply/distribute again.

`f(x)=3x^3+10x^2+3x-15x^2-50x-15`

Combine like terms.

`f(x)=3x^3-5x^2-47x-15`