How do you factor y= 6x^3+13x-14x+3 ?

Dec 30, 2015

Use the rational root theorem to help find the first root and factor, then an AC method to factor the remaining quadratic to find:

$6 {x}^{3} + 13 {x}^{2} - 14 x + 3 = \left(3 x - 1\right) \left(2 x - 1\right) \left(x + 3\right)$

Explanation:

I will guess that $13 x$ should have been $13 {x}^{2}$ in the question.

Let $f \left(x\right) = 6 {x}^{3} + 13 {x}^{2} - 14 x + 3$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in lowest terms in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $p$ a divisor of the constant term $3$ and $q$ a divisor of the coefficient $6$ of the leading term.

That means that the only possible rational roots are:

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

Let's try some of these:

$f \left(\frac{1}{6}\right) = \frac{6}{216} + \frac{13}{36} - \frac{14}{6} + 3$

$= \frac{1 + 13 - 84 + 108}{36} = \frac{38}{36} = \frac{19}{18}$

$f \left(- \frac{1}{6}\right) = \frac{- 1 + 13 + 84 + 108}{36} = \frac{204}{36} = \frac{17}{3}$

$f \left(\frac{1}{3}\right) = \frac{6}{27} + \frac{13}{9} - \frac{14}{3} + 3$

$= \frac{2 + 13 - 42 + 27}{9} = 0$

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

$6 {x}^{3} + 13 {x}^{2} - 14 x + 3 = \left(3 x - 1\right) \left(2 {x}^{2} + 5 x - 3\right)$

Then use an AC method to help factor $2 {x}^{2} + 5 x - 3$.

Look for a pair of factors of $A C = 2 \cdot 3 = 6$ that differ by $B = 5$.

The pair $6 , 1$ works, hence:

$2 {x}^{2} + 5 x - 3$

$= 2 {x}^{2} + 6 x - x - 3$

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

$= 2 x \left(x + 3\right) - 1 \left(x + 3\right)$

$= \left(2 x - 1\right) \left(x + 3\right)$

Putting it all together:

$6 {x}^{3} + 13 {x}^{2} - 14 x + 3 = \left(3 x - 1\right) \left(2 x - 1\right) \left(x + 3\right)$