# The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=3 and roots of multiplicity 1 at x=0 and x=-3. It goes through the point (5,112). How do you find a formula for P(x)?

Sep 26, 2017

A polynomial of degree 4 will have the root form:

$y = k \left(x - {r}_{1}\right) \left(x - {r}_{2}\right) \left(x - {r}_{3}\right) \left(x - {r}_{4}\right)$

Substitute in the values for the roots and then use the point to find the value of k.

#### Explanation:

Substitute in the values for the roots:

$y = k \left(x - 0\right) \left(x - 3\right) \left(x - 3\right) \left(x - \left(- 3\right)\right)$

Use the point $\left(5 , 112\right)$ to find the value of k:

$112 = k \left(5 - 0\right) \left(5 - 3\right) \left(5 - 3\right) \left(5 - \left(- 3\right)\right)$

$112 = k \left(5\right) \left(2\right) \left(2\right) \left(8\right)$

$k = \frac{112}{\left(5\right) \left(2\right) \left(2\right) \left(8\right)}$

$k = \frac{7}{10}$

The root from of the polynomial is:

$y = \frac{7}{10} \left(x - 0\right) \left(x - 3\right) \left(x - 3\right) \left(x - \left(- 3\right)\right)$