The polynomial of degree $$ 3, P(x) $$, has a root of multiplicity 2 at $$ x=5 $$ and a root of multiplicity 1 at $$ x=-2 $$. The $$ y $$-intercept is $$ y=-35 $$. Find a formula for $$ P(x) $$. $$ P(x)=\square $$

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Given that the polynomial $$ P(x) $$ of degree 3 has a root of multiplicity 2 at $$ x=5 $$ and a root of multiplicity 1 at $$ x=-2 $$, and the $$ y $$-intercept is $$ y=-35 $$, we can determine the formula for $$ P(x) $$ as follows:

1. Since the root of multiplicity 2 at $$ x=5 $$, the factor corresponding to this root is $$ (x-5)^2 $$.
2. Since the root of multiplicity 1 at $$ x=-2 $$, the factor corresponding to this root is $$ (x+2) $$.
3. The $$ y $$-intercept is given as $$ y=-35 $$, which means when $$ x=0 $$, $$ P(x) = -35 $$.

Therefore, the formula for $$ P(x) $$ can be expressed as:
$$ P(x) = a(x-5)^2(x+2) $$
where $$ a $$ is a constant that we need to determine.

To find the value of $$ a $$, we can substitute $$ x=0 $$ into the equation and solve for $$ a $$:
$$ P(0) = a(0-5)^2(0+2) = -35 $$
$$ a(-5)^2(2) = -35 $$
$$ 25a(2) = -35 $$
$$ 50a = -35 $$
$$ a = -\frac{35}{50} $$
$$ a = -\frac{7}{10} $$

Therefore, the formula for $$ P(x) $$ is:
$$ P(x) = -\frac{7}{10}(x-5)^2(x+2) $$
$$ P(x) = -\frac{7}{10}(x-5)^2(x+2) $$

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 .