# How do you list all possible roots and find all factors and zeroes of x^3+4x^2+5x+2?

##### 1 Answer
Sep 1, 2016

$p \left(x\right) = {x}^{3} + 4 {x}^{2} + 5 x + 2 = {\left(x + 1\right)}^{2} \left(x + 2\right)$.

The roots of $\text{p(x)=0, &, the zeroes of p(x) are, } - 1 , - 1 , - 2$.

#### Explanation:

Name the given poly. $p \left(x\right) = {x}^{3} + 4 {x}^{2} + 5 x + 2$.

We notice that,

"The sum of the co-effs. of odd-powered terms"=1+5=6, &,

$\text{The sum of the co-effs. of even-powered terms} = 4 + 2 = 6.$

This means that $x + 1$ is a factor of $p \left(x\right)$.

$\text{Now p(x)} = {x}^{3} + 4 {x}^{2} + 5 x + 2$,

$= \underline{{x}^{3} + {x}^{2}} + \underline{3 {x}^{2} + 3 x} + \underline{2 x + 2}$,

$= {x}^{2} \left(x + 1\right) + 3 x \left(x + 1\right) + 2 \left(x + 1\right)$,

$= \left(x + 1\right) \left({x}^{2} + 3 x + 2\right)$,

$= \left(x + 1\right) \left\{\underline{{x}^{2} + 2 x} + \underline{x + 1}\right\}$,

$= \left(x + 1\right) \left\{x \left(x + 1\right) + 1 \left(x + 1\right)\right\}$,

$= \left(x + 1\right) \left\{\left(x + 1\right) \left(x + 2\right)\right\}$,

$= {\left(x + 1\right)}^{2} \left(x + 2\right)$.

Clearly, the roots of $\text{p(x)=0, &, the zeroes of p(x) are, } - 1 , - 1 , - 2$.