# How do you create a polynomial p which has zeros x=+-3, x=6, leading term is 7x^5, and the point (-3,0) is a local minimum on the graph of y=p(x)?

Feb 19, 2018

$f \left(x\right) = 7 {x}^{5} + 21 {x}^{4} - 315 {x}^{3} - 945 {x}^{2} + 2268 x + 6804$

#### Explanation:

Each zero $x = a$ corresponds to a linear factor $\left(x - a\right)$

We need the zero at $\left(- 3 , 0\right)$ to be of even multiplicity in order that it is also a local minimum or maximum.

Also, since the leading coefficient is positive and the point $\left(- 3 , 0\right)$ is a local minimum, there must be another real zero less than $- 3$.

So let's write:

$f \left(x\right) = 7 \left(x + 6\right) \left(x + 3\right) \left(x + 3\right) \left(x - 3\right) \left(x - 6\right)$

$\textcolor{w h i t e}{f \left(x\right)} = 7 \left(x + 3\right) \left({x}^{2} - 36\right) \left({x}^{2} - 9\right)$

$\textcolor{w h i t e}{f \left(x\right)} = 7 {x}^{5} + 21 {x}^{4} - 315 {x}^{3} - 945 {x}^{2} + 2268 x + 6804$

graph{7x^5+21x^4-315x^3-945x^2+2268x+6804 [-10, 10, -12000, 10000]}