# A polynomial function p can be factored into seven factors: ( x-3), ( x+1), and 5 factors of (x -2). What are its zeros with multiplicity, and what is the degree of the polynomial? Explain.

Nov 29, 2017

#### Answer:

Zero 3 has multiplicity 1, zero -1 has multiplicity 1, and zero 2 has multiplicity 5. The degree of the polynomial is 7.

#### Explanation:

$p \left(x\right) = \left(x - 3\right) \left(x + 1\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right)$

The multiplicity of an equation is how many times a zero repeats.

$p \left(x\right) = \left(x - 3\right) \left(x + 1\right) {\left(x - 2\right)}^{5}$

In the equation, zero 3 has multiplicity 1, zero -1 has multiplicity 1, and zero 2 has multiplicity 5.

The degree of the whole polynomial is the highest degree out of every term.
So first you have to expand:
$p \left(x\right) = \left(x - 3\right) \left(x + 1\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right)$
$p \left(x\right) = \left({x}^{2} + x - 3 x - 3\right) \left({x}^{2} - 4 x + 4\right) \left({x}^{2} - 4 x + 4\right) \left(x - 2\right)$
$p \left(x\right) = \left({x}^{2} - 2 x - 3\right) \left({x}^{2} - 4 + 4\right) \left({x}^{3} - 6 {x}^{2} + 12 x - 8\right)$
$p \left(x\right) = \left({x}^{4} - 6 {x}^{3} + 9 {x}^{2} + 4 x - 12\right) \left({x}^{3} - 6 {x}^{2} + 12 x - 8\right)$
$p \left(x\right) = {x}^{7} - 12 {x}^{6} + 57 {x}^{5} - 130 {x}^{4} + 120 {x}^{3} + 48 {x}^{2} - 176 x + 96$

So the degree of $p \left(x\right) = \left(x - 3\right) \left(x + 1\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right) \left(x - 2\right)$ is 7.