The polynomial of degree $$ 3, P(x) $$, has a root of multiplicity 2 at $$ x=4 $$ and a root of multiplicity 1 at $$ x=-4 $$. The $$ y $$-intercept is $$ y=-19.2 $$. Find a formula for $$ P(x) $$. $$ P(x)= $$

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We are given that $$ P(x) $$ is a cubic polynomial with a double root at $$ x=4 $$ and a single root at $$ x=-4 $$. Thus, we can write

$$
P(x) = k(x-4)^2(x+4)
$$

where $$ k $$ is a constant.

To find $$ k $$, we use the $$ y $$-intercept. Since $$ P(0) = -19.2 $$, substitute $$ x=0 $$ into the polynomial:

$$
P(0) = k(0-4)^2(0+4) = k \cdot 16 \cdot 4 = 64k
$$

So

$$
64k = -19.2
$$

Solving for $$ k $$:

$$
k = \frac{-19.2}{64} = -0.3
$$

Thus, the polynomial is

$$
P(x) = -0.3(x-4)^2(x+4)
$$
$$ P(x) = -0.3(x - 4)^2(x + 4) $$

The polynomial of degree , has a root of multiplicity 2 at
and a root of multiplicity 1 at . The -intercept is
.
Find a formula for .

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The polynomial of degree , has a root of multiplicity 2 at and a root of multiplicity 1 at . The -intercept is . Find a formula for .