# What are all the possible rational zeros for f(x)=x^3+3x^2+x-2 and how do you find all zeros?

Jan 9, 2017

All the zeroes of $f \left(x\right)$ are,-2,(-1+-sqrt5)/2; among these,

only rational zero is $- 2$.

#### Explanation:

We observe that the sum of the co-effs. of

$f \left(x\right) \text{ is } 1 + 3 + 1 - 2 = 3 \ne 0 , \therefore \left(x - 1\right)$ is not a factor.

Also, the sum of the co-effs. of odd-powered terms is 2, and that

of even-powered 1, Since, these are not equal, so, $\left(x + 1\right)$ is also

not a factor.

Now, $\left(p x + q\right) | \left({x}^{3} + 3 {x}^{2} + x - 2\right) \Rightarrow p | 1 \mathmr{and} q | \left(- 2\right) .$

Hence, the possible linear factors $\left(p x + q\right) \text{ of } f \left(x\right)$ are,

$\left(\pm x \pm 1\right) , \left(\pm x \pm 2\right)$

We have already verified that $\left(\pm x \pm 1\right)$ are not factors.

For $\left(x + 2\right) , \text{we find, } f \left(- 2\right) = - 8 + 12 - 2 - 2 = 0$

$\therefore \left(x + 2\right) \text{ is a factor}$.

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

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

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

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

For the Quadratic ${x}^{2} + x - 1 , \Delta = {b}^{2} - 4 a c = 5.$

$\therefore \text{ its zeroes are, } \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 1 \pm \sqrt{5}}{2}$.

Thus, all the zeroes of $f \left(x\right)$ are,-2,(-1+-sqrt5)/2; among these,

only rational zero is $- 2$.

Enjoy Maths.!

Jan 9, 2017

All the possible rational zeros for f(x) are +1;-1;+2;-2;
all zeros are -2 (rational) and $\frac{- 1 \pm \sqrt{5}}{2}$ (not rational)

#### Explanation:

All the possible rational zeros for f(x) are the factors of the known term 2:

+1;-1;+2;-2

You would apply the remainder rule to find the first zero:

$\textcolor{red}{f \left(1\right) = 1 + 3 + 1 - 2 = 3 \ne 0}$

and you conclude that 1 isn't a zero for f(x);

$\textcolor{red}{f \left(- 1\right) = {\left(- 1\right)}^{3} + 3 {\left(- 1\right)}^{2} + \left(- 1\right) - 2 = - 1 + 3 - 1 - 2 = - 1 \ne 0}$

and you conclude that -1 isn't a zero for f(x);

$\textcolor{red}{f \left(2\right) = {2}^{3} + 3 \cdot {2}^{2} \cancel{+ 2} \cancel{- 2} = 8 + 12 = 20 \ne 0}$

and you conclude that 2 isn't a zero for f(x);

$\textcolor{g r e e n}{f \left(- 2\right) = {\left(- 2\right)}^{3} + 3 \cdot {\left(- 2\right)}^{2} + \left(- 2\right) - 2 = - 8 + 12 - 2 - 2 = 0}$

then -2 is a zero for f(x)

Then you would divide f(x) by (x+2) and get the polynomial:

${x}^{2} + x - 1$

that, solved by the quadratic formula, will give the not rational zeros:

$\frac{- 1 \pm \sqrt{5}}{2}$