# How do you find all the zeros of P(x) = (x-5)^2(x+2)^3(x+4)?

May 28, 2016

Note that $x$ is a zero if and only it causes at least one of the factors of $P \left(x\right)$ to be zero. Hence zeros:

$5$ (multiplicity $2$)

$- 2$ (multiplicity $3$)

$- 4$

#### Explanation:

$P \left(x\right) = {\left(x - 5\right)}^{2} {\left(x + 2\right)}^{3} \left(x + 4\right)$

$= \left(x - 5\right) \left(x - 5\right) \left(x + 2\right) \left(x + 2\right) \left(x + 2\right) \left(x + 4\right)$

Note that $P \left(x\right) = 0$ if and only if at least one of its factors is zero.

So the only zeros are:

• $x = 5$ with multiplicity $2$
• $x = - 2$ with multiplicity $3$
• $x = - 4$ with multiplicity $1$

Essentially, finding the zeros of a polynomial (with their multiplicities) is the same as finding the linear factors of that polynomial. If $\left(x - a\right)$ is a factor then $x = a$ is a zero and vice versa.