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)?

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Answer 1 · 2017-09-26T21:11:32

A polynomial of degree 4 will have the root form:

$y = k (x - r_{1}) (x - r_{2}) (x - r_{3}) (x - r_{4})$ $y=k(x-r_1)(x-r_2)(x-r_3)(x-r_4)$

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

Substitute in the values for the roots:

$y = k (x - 0) (x - 3) (x - 3) (x - (- 3))$ $y=k(x-0)(x-3)(x-3)(x-(-3))$

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

$112 = k (5 - 0) (5 - 3) (5 - 3) (5 - (- 3))$ $112=k(5-0)(5-3)(5-3)(5-(-3))$

$112 = k (5) (2) (2) (8)$ $112=k(5)(2)(2)(8)$

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

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

The root from of the polynomial is:

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