# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -5, 2, -2?

Feb 28, 2017

${x}^{3} + 5 {x}^{2} - 4 x - 20 = 0$

#### Explanation:

$\text{ if the zeros are" -5,2,-2, " we write down the corresponding factors and multiply them}$

zeros

$x = - 5 \implies \left(x + 5\right) = 0$

$x = 2 \implies \left(x - 2\right) = 0$

$x = - 2 \implies \left(x + 2\right) = 0$

the polynomial is:

$\left(x + 5\right) \left(x - 2\right) \left(x + 2\right) = 0$

multiplying out.

$\left(x + 5\right) \left({x}^{2} - 4\right) = 0 \text{ (difference of squares)}$

${x}^{3} - 4 x + 5 {x}^{2} - 20 = 0$

tidying up

${x}^{3} + 5 {x}^{2} - 4 x - 20 = 0$