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

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

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Answer 1 · 2015-12-30T19:09:24

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

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

Explanation:

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

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

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

That means that the only possible rational roots are:

$\pm \frac{1}{6}$ $+-1/6$ , $\pm \frac{1}{3}$ $+-1/3$ , $\pm \frac{1}{2}$ $+-1/2$ , $\pm 1$ $+-1$ , $\pm \frac{3}{2}$ $+-3/2$ , $\pm 3$ $+-3$

Let's try some of these:

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

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

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

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

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

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

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

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

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

The pair $6, 1$ $6, 1$ works, hence:

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

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

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

$= 2 x (x + 3) - 1 (x + 3)$ $= 2x(x+3)-1(x+3)$

$= (2 x - 1) (x + 3)$ $= (2x-1)(x+3)$

Putting it all together:

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

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

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How do you factor y=6x3+13x−14x+3y= 6x^3+13x-14x+3 ?

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