How do you completely factor $6 x^{3} + 7 x^{2} - 1$ $6x^3+7x^2-1$ given $2 x + 1$ $2x+1$ is one of the factors?

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How do you completely factor $6 x^{3} + 7 x^{2} - 1$ $6x^3+7x^2-1$ given $2 x + 1$ $2x+1$ is one of the factors?

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Answer 1 · 2016-10-22T14:23:37

The complete factorization is $(2 x + 1) (3 x - 1) (x + 1)$ $(2x+1)(3x-1)(x+1)$

Explanation:

We have to make long division

$6 x^{3} + 7 x^{2}$ $6x^3+7x^2$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a a a$ $color(white)(aaaaaa)$ $∣ 2 x + 1$ $∣2x+1$
$6 x^{3} + 3 x^{2}$ $6x^3+3x^2$ $a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaa)$ $3 x^{2} + 2 x - 1$ $3x^2+ 2x-1$
$0 + 4 x^{2} + 2 x$ $0 +4x^2 +2x$
$a a a a a a a$ $color(white)(aaaaaaa)$ $- 2 x - 1$ $-2x-1$
$a a a a a a a a$ $color(white)(aaaaaaaa)$ $+ 0 - 0$ $+0 -0$

So the result of the long division is $3 x^{2} + 2 x - 1$ $3x^2+2x-1$

which upon factorization gives $(3 x - 1) (x + 1)$ $(3x-1)(x+1)$

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How do you completely factor $6 x^{3} + 7 x^{2} - 1$ $6x^3+7x^2-1$ given $2 x + 1$ $2x+1$ is one of the factors?

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Explanation:

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