How do you list all possible roots and find all factors of $6 x^{3} + 7 x^{2} - 3 x - 1$ $6x^3+7x^2-3x-1$ ?

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How do you list all possible roots and find all factors of $6 x^{3} + 7 x^{2} - 3 x - 1$ $6x^3+7x^2-3x-1$ ?

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Answer 1 · 2017-01-17T00:52:30

The "possible" rational zeros are:

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

The factorisation is:

$6 x^{3} + 7 x^{2} - 3 x - 1$ $6x^3+7x^2-3x-1$

$= 3 (2 x - 1) (x + \frac{5}{6} - \frac{\sqrt{13}}{6}) (x + \frac{5}{6} + \frac{\sqrt{13}}{6})$ $= 3(2x-1)(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)$

Explanation:

Given:

$f (x) = 6 x^{3} + 7 x^{2} - 3 x - 1$ $f(x) = 6x^3+7x^2-3x-1$

By the rational roots theorem, any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $- 1$ $-1$ and $q$ $q$ a divisor of the coefficient $6$ $6$ of the leading term.

That means that the only possible rational zeros are:

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

We find:

$f (\frac{1}{2}) = 6 (\frac{1}{8}) + 7 (\frac{1}{4}) - 3 (\frac{1}{2}) - 1 = \frac{3 + 7 - 6 - 4}{4} = 0$ $f(1/2) = 6(1/8)+7(1/4)-3(1/2)-1 = (3+7-6-4)/4 = 0$

So $x = \frac{1}{2}$ $x=1/2$ is a zero and $(2 x - 1)$ $(2x-1)$ a factor:

$6 x^{3} + 7 x^{2} - 3 x - 1 = (2 x - 1) (3 x^{2} + 5 x + 1)$ $6x^3+7x^2-3x-1 = (2x-1)(3x^2+5x+1)$

We can factor the remaining quadratic by completing the square.

To reduce the need to do arithmetic with fractions, I will premultiply by $3 \cdot 2^{2} = 12$ $3*2^2 = 12$ and divide by it at the end:

$12 (3 x^{2} + 5 x + 1) = 36 x^{2} + 60 x + 12$ $12(3x^2+5x+1) = 36x^2+60x+12$

$12 (3 x^{2} + 5 x + 1) = {(6 x)}^{2} + 2 (6 x) (5) + 25 - 13$ $color(white)(12(3x^2+5x+1)) = (6x)^2+2(6x)(5)+25-13$

$12 (3 x^{2} + 5 x + 1) = {(6 x + 5)}^{2} - {(\sqrt{13})}^{2}$ $color(white)(12(3x^2+5x+1)) = (6x+5)^2-(sqrt(13))^2$

$12 (3 x^{2} + 5 x + 1) = ((6 x + 5) - \sqrt{13}) ((6 x + 5) + \sqrt{13})$ $color(white)(12(3x^2+5x+1)) = ((6x+5)-sqrt(13))((6x+5)+sqrt(13))$

$12 (3 x^{2} + 5 x + 1) = (6 x + 5 - \sqrt{13}) (6 x + 5 + \sqrt{13})$ $color(white)(12(3x^2+5x+1)) = (6x+5-sqrt(13))(6x+5+sqrt(13))$

$12 (3 x^{2} + 5 x + 1) = 12 \cdot 3 (x + \frac{5}{6} - \frac{\sqrt{13}}{6}) (x + \frac{5}{6} + \frac{\sqrt{13}}{6})$ $color(white)(12(3x^2+5x+1)) = 12*3(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)$

So:

$3 x^{2} + 60 x + 12 = 3 (x + \frac{5}{6} - \frac{\sqrt{13}}{6}) (x + \frac{5}{6} + \frac{\sqrt{13}}{6})$ $3x^2+60x+12 = 3(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)$

Putting it all together:

$6 x^{3} + 7 x^{2} - 3 x - 1$ $6x^3+7x^2-3x-1$

$= 3 (2 x - 1) (x + \frac{5}{6} - \frac{\sqrt{13}}{6}) (x + \frac{5}{6} + \frac{\sqrt{13}}{6})$ $= 3(2x-1)(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)$

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