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 $y=p(x)$ ?

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Answer 1 · 2018-02-19T19:52:00

$f(x) = 7x^5+21x^4-315x^3-945x^2+2268x+6804$

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

We need the zero at $(-3, 0)$ 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 $(-3, 0)$ is a local minimum, there must be another real zero less than $-3$ .

So let's write:

$f(x) = 7(x+6)(x+3)(x+3)(x-3)(x-6)$

$color(white)(f(x)) = 7(x+3)(x^2-36)(x^2-9)$

$color(white)(f(x)) =7x^5+21x^4-315x^3-945x^2+2268x+6804$

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

How do you create a polynomial pp which has zeros x=+-3, x=6x=+-3, x=6, leading term is 7x^57x^5, and the point (-3,0)(-3,0) is a local minimum on the graph of y=p(x)y=p(x)?