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

Sep 24, 2016

$y = {x}^{3} - 7 {x}^{2} + 14 x - 8$

#### Explanation:

If the zeros of a polynomial are $x = a , x = b \text{ and } x = c$

Then the factors of the polynomial are

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

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

here the zeros are $x = 2 , x = 1 \text{ and } x = 4$

$\Rightarrow \text{factors are " (x-2),(x-1)" and } \left(x - 4\right)$

$\Rightarrow y = \left(x - 2\right) \left(x - 1\right) \left(x - 4\right)$

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

$= {x}^{3} - 3 {x}^{2} + 2 x - 4 {x}^{2} + 12 x - 8$

$\Rightarrow y = {x}^{3} - 7 {x}^{2} + 14 x - 8 \text{ is the polynomial}$