# What is a cubic polynomial function in standard form with zeros 3, -4, and 5?

Feb 16, 2018

$y = {x}^{3} - 4 {x}^{2} - 17 x + 60$

#### Explanation:

.

The standard form of a cubic polynomial function is:

$y = a {x}^{3} + b {x}^{2} + c x + d$

If we know the zeros as being $\alpha$, $\beta$, and $\theta$ then we know its factored form to be:

$y = \left(x - \alpha\right) \left(x - \beta\right) \left(x - \theta\right)$

simply because you would have had to set the expression within each set of parentheses equal to $0$ and solve for a root.

Therefore, we can write the polynomial in this problem as:

$y = \left(x - 3\right) \left(x + 4\right) \left(x - 5\right)$

Now, if we multiply these through and remove the parentheses we will have the standard form of it:

$y = \left(x - 3\right) \left({x}^{2} - x - 20\right)$

$y = {x}^{3} - {x}^{2} - 20 x - 3 {x}^{2} + 3 x + 60$

$y = {x}^{3} - 4 {x}^{2} - 17 x + 60$

Feb 16, 2018

${x}^{3} - 4 {x}^{2} - 17 x + 60$

#### Explanation:

There is only one cubic polynomial in standard form (i.e. monic) with three given distinct zeros, $a , b$ and $c$.
It is given by the product:

$\left(x - a\right) \left(x - b\right) \left(x - c\right)$

$\left(x - 3\right) \left(x + 4\right) \left(x - 5\right) = \left({x}^{2} + x - 12\right) \left(x - 5\right)$
$= {x}^{3} - 4 {x}^{2} - 17 x + 60$