# How do you find a polynomial function that has zeros x=-4, -1, 3, 6 and degree n=4?

Jan 25, 2017

The Reqd. Poly. Fun. is

$k \left({x}^{4} - 4 {x}^{3} - 23 {x}^{2} + 54 x + 2\right) , k \in \mathbb{R} - \left\{0\right\} .$

#### Explanation:

Let us denote by $p \left(x\right)$ the reqd. poly. fun.

$x = - 4 \text{ is a zero of } p \left(x\right) \therefore \left(x - \left(- 4\right)\right) = \left(x + 4\right)$ is a factor.

On the similar lines, $\left(x + 1\right) , \left(x - 3\right) \mathmr{and} \left(x - 6\right)$ are also factors.

As the degree of $p \left(x\right)$ is $4 ,$ $p \left(x\right)$ can not have any more

factors, except some constant, say, $k \ne 0$.

Accordingly, we have,

$p \left(x\right) = k \left(x + 4\right) \left(x + 1\right) \left(x - 3\right) \left(x - 6\right)$

$= k \left\{\left(x + 4\right) \left(x - 6\right)\right\} \left(x + 1\right) \left(x - 3\right)$

$= k \left\{\left({x}^{2} - 2 x - 24\right) \left({x}^{2} - 2 x - 3\right)\right\}$

$= k \left(y - 24\right) \left(y - 3\right) , \left[y = {x}^{2} - 2 x\right]$

$= k \left({y}^{2} - 27 y + 72\right)$

$= k \left\{{\left({x}^{2} - 2 x\right)}^{2} - 27 \left({x}^{2} - 2 x\right) + 72\right\}$

$= k \left\{{x}^{4} - 4 {x}^{3} + 4 {x}^{2} - 27 {x}^{2} + 54 x + 72\right\}$

$\therefore p \left(x\right) = k \left({x}^{4} - 4 {x}^{3} - 23 {x}^{2} + 54 x + 2\right) , k \in \mathbb{R} - \left\{0\right\} ,$ is the reqd. poly.

Enjoy Maths.!