# How do you use the rational roots theorem to find all possible zeros of f(x)=3x^5-2x^4-15x^3+10x^2+12x-8 ?

Mar 3, 2016

Zeros of $f \left(x\right)$ are $\left\{- 2 , - 1 , 1 , \frac{2}{3.} 2\right\}$

#### Explanation:

Rational root theorem states that if a polynomial ${a}_{0} {x}^{n} + {a}_{1} {x}^{n - 1} + {a}_{2} {x}^{n - 2} + \ldots + {a}_{n}$, has rational roots $\frac{p}{q}$, where $p$ and $q$ are integers, then $q$ is a factor of ${a}_{0}$ and $p$ is a factor of ${a}_{n}$.

Hence to find all possible zeros of f(x)=3x^5−2x^4−15x^3+10x^2+12x−8,

we have to find roots of the equation 3x^5−2x^4−15x^3+10x^2+12x−8=0

For this we start by factors of constant term $- 8$ such as $\left\{1 , - 1 , 2 , - 2 , 4 , - 4 , 8 , - 8\right\}$.

It is observed that $x = 1$ makes $f \left(x\right) = 0$ and hence $\left(x - 1\right)$ is a factor of $f \left(x\right)$ as $f \left(1\right) = 3 - 2 - 15 + 10 + 12 - 8 = 0$.

Similarly for $x = - 1$ makes $f \left(x\right) = 0$ and hence $\left(x + 1\right)$ is a factor of $f \left(x\right)$ as $f \left(1\right) = - 3 - 2 + 15 + 10 - 12 - 8 = 0$.

$x = - 2$ makes $f \left(x\right) = 0$ as $f \left(2\right) = 3 \cdot 32 - 2 \cdot 16 - 15 \cdot 8 + 10 \cdot 4 + 12 \cdot 2 - 8 = 96 - 32 - 120 + 40 + 24 - 8 = 0$ and hence $\left(x - 2\right)$ is a factor of $f \left(x\right)$.

$x = - - 2$ makes $f \left(x\right) = 0$ as $f \left(2\right) = 3 \cdot \left(- 32\right) - 2 \cdot 16 - 15 \cdot \left(- 8\right) + 10 \cdot 4 + 12 \cdot \left(- 2\right) - 8 = - 96 - 32 + 120 + 40 - 24 - 8 = 0$ and hence $\left(x + 2\right)$ is a factor of $f \left(x\right)$.

As four factors of $f \left(x\right)$ have been found, we can have fifth factor by dividing $f \left(x\right)$ by these and result would be $\left(3 x - 2\right)$ and hence $\left(3 x - 2\right) = 0$ or $x = \frac{2}{3}$ is another zero of $f \left(x\right)$

Hence, zeros of $f \left(x\right)$ are $\left\{- 2 , - 1 , 1 , \frac{2}{3.} 2\right\}$