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

May 4, 2016

$x = - 2 , 1 , - 1$

#### Explanation:

So, I like to factor this sort of problm using synthetic division. First thing's first, let's set up this problem.

$\textcolor{w h i t e}{- 2} |$$1 \textcolor{w h i t e}{000} 2 \textcolor{w h i t e}{000} - 1 \textcolor{w h i t e}{000} - 2$
$\textcolor{w h i t e}{- 2} |$
$- 2 |$__________

We bring down the $1$, which gives us this:

$\textcolor{w h i t e}{- 2} |$$1 \textcolor{w h i t e}{000} 2 \textcolor{w h i t e}{000} - 1 \textcolor{w h i t e}{000} - 2$
$\textcolor{w h i t e}{- 2} |$
$- 2 |$__________
$\textcolor{w h i t e}{0} \textcolor{w h i t e}{000} 1$

We then multiply the $- 2$ by the $1$ that we brought down to give us $- 2$. That value is brought up to the next row, which brings us to this:

$\textcolor{w h i t e}{- 2} |$$1 \textcolor{w h i t e}{000} 2 \textcolor{w h i t e}{000} - 1 \textcolor{w h i t e}{000} - 2$
$\textcolor{w h i t e}{- 2} |$$\textcolor{w h i t e}{00} - 2$
$- 2 |$__________
$\textcolor{w h i t e}{0} \textcolor{w h i t e}{000} 1$

From here, we just add the $2$ to the $- 2$ and end up at this:

$\textcolor{w h i t e}{- 2} |$$1 \textcolor{w h i t e}{000} 2 \textcolor{w h i t e}{000} - 1 \textcolor{w h i t e}{000} - 2$
$\textcolor{w h i t e}{- 2} |$$\textcolor{w h i t e}{00} - 2$
$- 2 |$__________
$\textcolor{w h i t e}{0} \textcolor{w h i t e}{000} 1 \textcolor{w h i t e}{000} 0$

If we continue this system, we end up with this:

$\textcolor{w h i t e}{- 2} |$$1 \textcolor{w h i t e}{000} 2 \textcolor{w h i t e}{000} - 1 \textcolor{w h i t e}{000} - 2$
$\textcolor{w h i t e}{- 2} |$$\textcolor{w h i t e}{00} - 2 \textcolor{w h i t e}{0000} 0 \textcolor{w h i t e}{000000} 2$
$- 2 |$________
$\textcolor{w h i t e}{0} \textcolor{w h i t e}{000} 1 \textcolor{w h i t e}{000} 0 \textcolor{w h i t e}{00} - 1 \textcolor{w h i t e}{000000} 0$

So now we have $x = - 2$ or $x + 2 = 0$, or $\left(x + 2\right)$ and $\left({x}^{2} - 1\right)$. This can be simplified to just $\left(x + 1\right) \left(x - 1\right)$.

We now have the factors of ${x}^{3} + 2 {x}^{2} - x - 2$, which are $\left(x + 2\right) \left(x + 1\right) \left(x - 1\right)$, which can be rewritten as: $x = - 2 , - 1 , 1$.

So let's just double check that we found all the zeros of ${x}^{3} + 2 {x}^{2} - x - 2$ be graphing it

graph{x^3+2x^2-x-2}

And look, the three $x$-intercepts are what we said, $x = 1 , - 1 , - 2$.

Jun 14, 2016

The zeros are $- 2 , - 1 , 1.$

#### Explanation:

Method 1

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

Hence, the zeros are $- 2 , - 1 , 1.$

Method 2

Observe that the sum of the co-effs. $= 1 + 2 - 1 - 2 = 0$

So, $\left(x - 1\right)$ is a factor of poly. $f \left(x\right)$. Now, we will rearrange the terms of the poly. in such a way that $\left(x - 1\right)$ can be taken out as a common factor. See the following computation :-

$f \left(x\right) = {x}^{3} + \textcolor{red}{2} {x}^{2} - \textcolor{b l u e}{1} x - 2 = {x}^{3} - \textcolor{red}{1} {x}^{2} + \textcolor{red}{3} {x}^{2} - \textcolor{b l u e}{3} x + \textcolor{b l u e}{2} x - 2 = {x}^{2} \left(x - 1\right) + 3 x \left(x - 1\right) + 2 \left(x - 1\right) = \left(x - 1\right) \left({x}^{2} + 3 x + 2\right) = \left(x - 1\right) \left\{{x}^{2} + \left(2 + 1\right) x + \left(2 \times 1\right)\right\} = \left(x - 1\right) \left\{{x}^{2} + 2 x + x + 2\right\} = \left(x - 1\right) \left\{x \left(x + 2\right) + 1 \left(x + 2\right)\right\} = \left(x - 1\right) \left(x + 2\right) \left(x + 1\right) .$

Hence, the zeros are $1 , - 2 , - 1$ as before!