Question

How do you write a polynomial in standard form given the zeros x=-6, 2, and 5?

Answer 1

A polynomial with zeros `x=-6,2,5` is `x^3-x^2-32x+60=0` in standard form.

Explanation:

You are given the following information about the polynomial: zeros.
Definition of zeros: If x = zero value, the polynomial becomes zero. i.e. If you plug in -6, 2, or 5 to x, this polynomial you are trying to find becomes zero.

For the polynomial to become zero at let's say x = 1, the polynomial needs to contain the following term:
`(x-1) = 0`
This looks too simple, and polynomials are usually bigger. They are usually quadratic, cubic, quartic, etc.
However, those messy polynomials all have `(x-1)=0` hidden in them if one of their zeros is 1.

Because we have 3 zeros, we write the same kind of equation for each zero.
`(x+6) = 0`
`(x-2) = 0`
`(x-5) = 0`

You multiply all of them to get the following equation.
`(x+6)(x-2)(x-5)=0`
If you are wondering why we multiply them, I'll give you a hint. Zero multiplied by any number is zero. So if one of the three equations above becomes zero, does it matter what values the other two equations give?

Expand the equation:
`(x^2+4x-12)(x-5)=0`
`x^3-x^2-32x+60=0`

And there's your answer!

One extra note: a more complete answer would be `a(x^3-x^2-32x+60=0)`. If this is homework, the `a` is probably not necessary as an answer. I am just telling you that you can multiply the whole polynomial by some number and not affect the zeros.

Answer 2

The polynomial equation with roots a, b and c is `x^3-(a+b+c)x^2+(bc+ca+ab)x-abc=0`. Answer: `x^3-13x^2+52x-60=0`-

Explanation:

The cubic equation with roots a, b and c is
#(x-a)(x-b)(x-c) =x^3-(a+b+c)x^2+(bc+ca+ab)x-abc=0#

Here (a, b, c) = (6, 2, 5).
a+b+c+ 13, bc+ca+ab=52 abd abc=60