# How do you use the rational root theorem to list all possible roots for 5x^3-11x^2+7x-1=0?

Nov 21, 2016

The "possible" rational roots are: $\pm \frac{1}{5} , \pm 1$

The actual zeros are: $1 , 1 , - \frac{1}{5}$

#### Explanation:

$f \left(x\right) = 5 {x}^{3} - 11 {x}^{2} + 7 x - 1$

By the rational root theorem, the only possible 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 $- 1$ and $q$ a divisor of the coefficient $5$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{5}$, $\pm 1$

There could be other zeros, but they will be irrational or non-Real Complex. The rational root theorem only tells us possible rational zeros.

Note that the sum of the coefficients of $f \left(x\right)$ is $0$. That is:

$5 - 11 + 7 - 1 = 0$

So $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$5 {x}^{3} - 11 {x}^{2} + 7 x - 1 = \left(x - 1\right) \left(5 {x}^{2} - 6 x + 1\right)$

The same is true of the remaining quadratic:

$5 - 6 + 1 = 0$

So we have another zero $x = 1$ and factor $\left(x - 1\right)$:

$5 {x}^{2} - 6 x + 1 = \left(x - 1\right) \left(5 x + 1\right)$

The remaining factor $\left(5 x + 1\right)$ corresponds to zero $x = - \frac{1}{5}$.