# Is it possible to factor y=6x^4-11x^3-51x^2+98x-27 ? If so, what are the factors?

Sep 19, 2016

$6 {x}^{4} - 11 {x}^{3} - 51 {x}^{2} + \textcolor{red}{99} x - 27 = \left(3 x - 1\right) \left(x - 3\right) \left(x + 3\right) \left(2 x - 3\right)$

#### Explanation:

I suspect a typo in the question.

Consider the quartic:

$f \left(x\right) = 6 {x}^{4} - 11 {x}^{3} - 51 {x}^{2} + \textcolor{red}{99} x - 27$

By the rational roots 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 $- 27$ and $q$ a divisor of the coefficient $6$ of the leading term.

That means that the only possible rational zeros of $f \left(x\right)$ are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3 , \pm \frac{9}{2} , \pm 9 , \pm \frac{27}{2} , \pm 27$

Trying each in turn, the first zero we come across is:

$f \left(\frac{1}{3}\right) = 6 \left(\frac{1}{81}\right) - 11 \left(\frac{1}{27}\right) - 51 \left(\frac{1}{9}\right) + 99 \left(\frac{1}{3}\right) - 27 = \frac{2 - 11 - 153 + 891 - 729}{27} = 0$

So $x = \frac{1}{3}$ is a zero and $\left(3 x - 1\right)$ a factor:

$6 {x}^{4} - 11 {x}^{3} - 51 {x}^{2} + 99 x - 27 = \left(3 x - 1\right) \left(2 {x}^{3} - 3 {x}^{2} - 18 x + 27\right)$

Notice that in the remaining cubic, the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

$2 {x}^{3} - 3 {x}^{2} - 18 x + 27 = \left(2 {x}^{3} - 3 {x}^{2}\right) - \left(18 x - 27\right)$

$\textcolor{w h i t e}{2 {x}^{3} - 3 {x}^{2} - 18 x + 27} = {x}^{2} \left(2 x - 3\right) - 9 \left(2 x - 3\right)$

$\textcolor{w h i t e}{2 {x}^{3} - 3 {x}^{2} - 18 x + 27} = \left({x}^{2} - 9\right) \left(2 x - 3\right)$

$\textcolor{w h i t e}{2 {x}^{3} - 3 {x}^{2} - 18 x + 27} = \left({x}^{2} - {3}^{2}\right) \left(2 x - 3\right)$

$\textcolor{w h i t e}{2 {x}^{3} - 3 {x}^{2} - 18 x + 27} = \left(x - 3\right) \left(x + 3\right) \left(2 x - 3\right)$

So we find:

$6 {x}^{4} - 11 {x}^{3} - 51 {x}^{2} + 99 x - 27 = \left(3 x - 1\right) \left(x - 3\right) \left(x + 3\right) \left(2 x - 3\right)$

$\textcolor{w h i t e}{}$
Footnote

The quartic as specified "will" factor, but only with very complicated coefficients. Here's one of them:

$\frac{11}{24} - \frac{1}{24} \sqrt{937 + \frac{10376 {3}^{\frac{2}{3}}}{\frac{1}{2} \left(37557 + i \sqrt{24771372627}\right)} ^ \left(\frac{1}{3}\right) + 4 {2}^{\frac{2}{3}} {\left(3 \left(37557 + i \sqrt{24771372627}\right)\right)}^{\frac{1}{3}}} - \frac{1}{2} \sqrt{\frac{937}{72} - {\left(\frac{1}{2} \left(37557 + i \sqrt{24771372627}\right)\right)}^{\frac{1}{3}} / \left(6 {3}^{\frac{2}{3}}\right) - \frac{1297}{3 {2}^{\frac{2}{3}} {\left(3 \left(37557 + i \sqrt{24771372627}\right)\right)}^{\frac{1}{3}}} + \frac{13429}{72 \sqrt{937 + \frac{10376 {3}^{\frac{2}{3}}}{\frac{1}{2} \left(37557 + i \sqrt{24771372627}\right)} ^ \left(\frac{1}{3}\right) + 4 {2}^{\frac{2}{3}} {\left(3 \left(37557 + i \sqrt{24771372627}\right)\right)}^{\frac{1}{3}}}}}$