# How do you create a polynomial p which has degree 4, as x->oo, p(x)->-oo, p has exactly three x intercepts (-6,0), (1,0) and (117,0), the graph of y=p(x) crosses through the x axis at (1,0)?

Apr 2, 2017

There are basically two possibilities:

$\left\{\begin{matrix}p \left(x\right) = k {\left(x + 6\right)}^{2} \left(x - 1\right) \left(x - 117\right) \\ p \left(x\right) = k \left(x + 6\right) \left(x - 1\right) {\left(x - 117\right)}^{2}\end{matrix}\right.$

with $k < 0$

#### Explanation:

The leading coefficient of $p \left(x\right)$ must be negative in order that $p \left(x\right) \to - \infty$ as $x \to \infty$.

It must have linear factors $\left(x + 6\right)$, $\left(x - 1\right)$ and $\left(x - 117\right)$

As there are only three $x$ intercepts, one of these must be repeated.

Also since $p \left(x\right)$ crosses the $x$ axis at $\left(1 , 0\right)$, the corresponding factor $\left(x - 1\right)$ only occurs an odd number of times, and thus exactly once.

So there are basically two possibilities:

$\left\{\begin{matrix}p \left(x\right) = k {\left(x + 6\right)}^{2} \left(x - 1\right) \left(x - 117\right) \\ p \left(x\right) = k \left(x + 6\right) \left(x - 1\right) {\left(x - 117\right)}^{2}\end{matrix}\right.$

with $k < 0$