# How do you write a polynomial function given the real zeroes 1,-4, 5 and coefficient 1?

Dec 6, 2015

The simplest such polynomial is:

$f \left(x\right) = \left(x - 1\right) \left(x + 4\right) \left(x - 5\right) = {x}^{3} - 2 {x}^{2} - 19 x + 20$

#### Explanation:

Convert the zeros into linear factors to find:

$f \left(x\right) = \left(x - 1\right) \left(x + 4\right) \left(x - 5\right) = {x}^{3} - 2 {x}^{2} - 19 x + 20$

Any polynomial in $x$ with these zeros will be a multiple (scalar or polynomial) of this $f \left(x\right)$.

Dec 6, 2015

Begin from the factored form to find the desired polynomial to be
$f \left(x\right) = {x}^{3} - 2 {x}^{2} - 19 x + 20$

#### Explanation:

An easy way of generating a polynomial with a given set of zeros is to begin with the factored form. A polynomial $f \left(x\right)$ has $a$ as a zero if and only if $\left(x - a\right)$ is a factor of $f \left(x\right)$.

Using this, we can construct the desired polynomial as follows:

$f \left(x\right) = \left(x - 1\right) \left(x - \left(- 4\right)\right) \left(x - 5\right)$

$= \left(x - 1\right) \left(x + 4\right) \left(x - 5\right)$

$= {x}^{3} - 2 {x}^{2} - 19 x + 20$

Note that we could multiply by a constant to give ${x}^{3}$ a different coefficient, however this method naturally produces the coefficient of the highest power as $1$.