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

Apr 3, 2016

A polynomial with zeros $x = - 6 , 2 , 5$ is ${x}^{3} - {x}^{2} - 32 x + 60 = 0$ 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:
$\left(x - 1\right) = 0$
This looks too simple, and polynomials are usually bigger. They are usually quadratic, cubic, quartic, etc.
However, those messy polynomials all have $\left(x - 1\right) = 0$ 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.
$\left(x + 6\right) = 0$
$\left(x - 2\right) = 0$
$\left(x - 5\right) = 0$

You multiply all of them to get the following equation.
$\left(x + 6\right) \left(x - 2\right) \left(x - 5\right) = 0$
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:
$\left({x}^{2} + 4 x - 12\right) \left(x - 5\right) = 0$
${x}^{3} - {x}^{2} - 32 x + 60 = 0$

One extra note: a more complete answer would be $a \left({x}^{3} - {x}^{2} - 32 x + 60 = 0\right)$. 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.

Apr 3, 2016

The polynomial equation with roots a, b and c is ${x}^{3} - \left(a + b + c\right) {x}^{2} + \left(b c + c a + a b\right) x - a b c = 0$. Answer: ${x}^{3} - 13 {x}^{2} + 52 x - 60 = 0$-

#### 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