# How do you list all possible roots and find all factors and zeroes of 4x^3-9x^2+6x-1?

Jul 3, 2016

$4 {x}^{3} - 9 {x}^{2} + 6 x - 1 = \left(4 x - 1\right) \left(x - 1\right) \left(x - 1\right)$

with zeros $x = 1$ with multiplicity $2$ and $x = \frac{1}{4}$.

#### Explanation:

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$f \left(x\right) = 4 {x}^{3} - 9 {x}^{2} + 6 x - 1$

I notice that the question asks for possible roots, so you are probably expected to make use of the rational root theorem first:

Since this cubic is given in standard form (with descending powers of $x$) and has integer coefficients, the rational root theorem can be applied:

Any rational zeros of $4 {x}^{3} - 9 {x}^{2} + 6 x - 1$ must be 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 $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1$

If we evaluate $f \left(\frac{1}{4}\right)$ we find:

$f \left(\frac{1}{4}\right) = \frac{4}{64} - \frac{9}{16} + \frac{6}{4} - 1 = \frac{1 - 9 + 24 - 16}{16} = 0$

So $x = \frac{1}{4}$ is a zero and $\left(4 x - 1\right)$ is a factor:

$4 {x}^{3} - 9 {x}^{2} + 6 x - 1$

$= \left(4 x - 1\right) \left({x}^{2} - 2 x + 1\right)$

$= \left(4 x - 1\right) {\left(x - 1\right)}^{2}$

Hence we have zeros:

$x = \frac{1}{4}$

$x = 1$ with multiplicity $2$

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Footnote

If the question did not mention "possible" roots, then I would have found the solution by looking at the sum of the coefficients first:

Note that $4 - 9 + 6 - 1 = 0$. Hence $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$4 {x}^{3} - 9 {x}^{2} + 6 x - 1 = \left(x - 1\right) \left(4 {x}^{2} - 5 x + 1\right)$

Then note that $4 - 5 + 1 = 0$. Hence $\left(x - 1\right)$ is a factor again:

$4 {x}^{2} - 5 x + 1 = \left(x - 1\right) \left(4 x - 1\right)$

Putting it together:

$4 {x}^{3} - 9 {x}^{2} + 6 x - 1 = \left(x - 1\right) \left(x - 1\right) \left(4 x - 1\right)$