The polynomial of degree 4, P(x) has a root multiplicity 2 at x=4 and roots multiplicity 1 at x=0 and x=-4 and it goes through the point (5, 18) how do you find a formula for p(x)?

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Answer 1 · 2016-10-27T05:40:12

The polynomial is $P (x) = \frac{2}{5} x {(x - 4)}^{2} (x + 4)$ $P(x)=2/5x(x-4)^2(x+4)$

If the polynomial has a root of multiplicity 2 at $x = 4$ $x=4$ , the ${(x - 4)}^{2}$ $(x-4)^2$
is a factor

Multiplicity 1 at $x = 0$ $x=0$ , then $x$ $x$ is a factor

Multiplicity 1 at $x = - 4$ $x=-4$ , then $(x + 4)$ $(x+4)$ is a factor

So $P (x) = A x {(x - 4)}^{2} (x + 4)$ $P(x)=Ax(x-4)^2(x+4)$

As it pases through $(5, 18)$ $(5,18)$ so
$18 = A \cdot 5 \cdot {(5 - 4)}^{2} \cdot (5 + 4)$ $18=A*5*(5-4)^2*(5+4)$

So $A = \frac{18}{5} \cdot \frac{1}{9} = \frac{2}{5}$ $A=18/5*1/9=2/5$

The polynomial is $P (x) = \frac{2}{5} x {(x - 4)}^{2} (x + 4)$ $P(x)=2/5x(x-4)^2(x+4)$