# What are all the possible rational zeros for y=x^4-2x^3-21x^2+22x+40 and how do you find all zeros?

##### 2 Answers
Nov 1, 2016

All the zeros are for $x = - 1 , 2 , - 4 , 5$

#### Explanation:

Let $f \left(x\right) = {x}^{4} - 2 {x}^{3} - 21 {x}^{2} + 22 x + 40$
By trial and error
$f \left(- 1\right) = 1 + 2 - 21 - 22 + 40 = 0$
And $f \left(2\right) = 16 - 16 - 84 + 44 + 40 = 0$
Therefore, $\left(x + 1\right) \left(x - 2\right)$ is a factor
We have tto make a long division
${x}^{4} - 2 {x}^{3} - 21 {x}^{2} + 22 x + 40$$\textcolor{w h i t e}{a a a a}$∣${x}^{2} - x - 2$
${x}^{4} - {x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$∣${x}^{2} - x - 20$
$\textcolor{w h i t e}{a}$$0 - {x}^{3} - 19 {x}^{2} + 22 x$
$\textcolor{w h i t e}{a a a}$$- {x}^{3} + {x}^{2} + 2 x$
$\textcolor{w h i t e}{a a a a}$$0 - 20 {x}^{2} + 20 x + 40$
$\textcolor{w h i t e}{a a a a a a}$$- 20 {x}^{2} + 20 x + 40$
So ${x}^{2} - x - 20 = \left(x + 4\right) \left(x - 5\right)$ these are factors
all the zeros are for $x = - 1 , 2 , - 4 , 5$

Nov 1, 2016

The "possible" rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 8 , \pm 10 , \pm 20 , \pm 40$

The actual zeros are: $- 1$, $2$, $5$ and $- 4$.

#### Explanation:

$\textcolor{w h i t e}{}$
$f \left(x\right) = {x}^{4} - 2 {x}^{3} - 21 {x}^{2} + 22 x + 40$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $40$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 8 , \pm 10 , \pm 20 , \pm 40$

Start by trying each in turn:

$f \left(1\right) = 1 - 2 - 21 + 22 + 40 = 40$

$f \left(- 1\right) = 1 + 2 - 21 - 22 + 40 = 0$

So $\textcolor{b l u e}{x = - 1}$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{4} - 2 {x}^{3} - 21 {x}^{2} + 22 x + 40 = \left(x + 1\right) \left({x}^{3} - 3 {x}^{2} - 18 x + 40\right)$

Let $g \left(x\right) = {x}^{3} - 3 {x}^{2} - 18 x + 40$

$g \left(- 1\right) = - 1 - 3 + 18 + 40 = 54$

$g \left(2\right) = {\left(2\right)}^{3} - 3 {\left(2\right)}^{2} - 18 \left(2\right) + 40 = 8 - 12 - 36 + 40 = 0$

So $\textcolor{b l u e}{x = 2}$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{3} - 3 {x}^{2} - 18 x + 40 = \left(x - 2\right) \left({x}^{2} - x - 20\right)$

To factor the remaining quadratic note, find a pair of factors of $20$ which differ by $1$. The pair $5 , 4$ works, so we find:

${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$

Hence the other two zeros are $\textcolor{b l u e}{x = 5}$ and $\textcolor{b l u e}{x = - 4}$.