# How do you write a polynomial with zeros -2, -4, -7 and leading coefficient 1?

Oct 3, 2016

$y = {x}^{3} + 13 {x}^{2} + 50 x + 56$

#### Explanation:

Given the zeros of a polynomial are $x = a , x = b , x = c$

Then its factors are $\left(x - a\right) , \left(x - b\right) \text{ and } \left(x - c\right)$

and $y = \left(x - a\right) \left(x - b\right) \left(x - c\right) \leftarrow \text{ the polynomial}$

here the zeros are $x = - 2 , x = - 4 \text{ and } x = - 7$

$\Rightarrow \left(x - \left(- 2\right)\right) , \left(x - \left(- 4\right)\right) , \left(x - \left(- 7\right)\right)$

That is $\left(x + 2\right) , \left(x + 4\right) , \left(x + 7\right) \text{ are the factors}$

$y = \left(x + 2\right) \left(x + 4\right) \left(x + 7\right) \text{ is the polynomial}$

distribute brackets to obtain the polynomial in standard form.

$y = \left(x + 2\right) \left({x}^{2} + 11 x + 28\right)$

$= {x}^{3} + 11 {x}^{2} + 28 x + 2 {x}^{2} + 22 x + 56$

$y = {x}^{3} + 13 {x}^{2} + 50 x + 56 \leftarrow \text{ is the polynomial}$