# What are all the possible rational zeros for f(x)=5x^3-11x^2+7x-1 and how do you find all zeros?

Sep 3, 2017

Zeros are $x = 0.2 \mathmr{and} x = 1$

#### Explanation:

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

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

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

= (x-1){5x(x-1) -1(x-1} = (x-1)(x-1)( 5x-1)

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

Zeros are

(x-1=0 or x=1 and 5x-1=0 or 5x=1 or x=0.2

graph{5x^3-11x^2+7x-1 [-10, 10, -5, 5]}
[Ans]

Sep 3, 2017

The "possible" rational zeros are:

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

The actual zeros are:

$1 \text{ }$ with multiplicity $2$

$\frac{1}{5} \text{ }$ with multiplicity $1$

#### Explanation:

Given:

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

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

That means that the only possible rational zeros are:

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

To find the actual zeros, we could just try each of these in turn, but there is a shortcut:

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

$5 - 11 + 7 - 1 = 0$

Hence $x = 1$ 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:

$5 - 6 + 1 = 0$

Hence $x = 1$ is a zero again and $\left(x - 1\right)$ a factor:

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

The remaining linear factor $5 x - 1$ gives us a zero $x = \frac{1}{5}$

So the zeros of $f \left(x\right)$ are:

$1 \text{ }$ with multiplicity $2$

$\frac{1}{5} \text{ }$ with multiplicity $1$

graph{5x^3-11x^2+7x-1 [-2.107, 2.89, -1.14, 1.36]}