Is it possible to factor $y = 4 x^{3} - 13 x - 6$ $y=4x^3-13x-6$ ? If so, what are the factors?

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Is it possible to factor $y = 4 x^{3} - 13 x - 6$ $y=4x^3-13x-6$ ? If so, what are the factors?

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Answer 1 · 2016-03-14T11:26:30

$4 x^{3} - 13 x - 6 = (x - 2) (2 x + 3) (2 x + 1)$ $4x^3-13x-6=(x-2)(2x+3)(2x+1)$

Explanation:

Since there is nothing obvious to factor out at the beginning, start by plugging in small numbers for $x$ $x$ to see if you can get the equation to equal $0$ $0$ .

Plugging in $x = 2$ $x=2$ gives $0$ $0$ , so $x - 2$ $x-2$ is a factor.

Now divide $4 x^{3} - 13 x - 6$ $4x^3-13x-6$ by $x - 2$ $x-2$ using polynomial long division or synthetic division. If you need help with this step, say something.

$\frac{4 x^{3} - 13 x - 6}{x - 2} = 4 x^{2} + 8 x + 3$ $(4x^3-13x-6)/(x-2)=4x^2+8x+3$

So, from here we know that

$4 x^{3} - 13 x - 6 = (x - 2) (4 x^{2} + 8 x + 3)$ $4x^3-13x-6=(x-2)(4x^2+8x+3)$

So now all we need to do is factor $4 x^{2} + 8 x + 3$ $4x^2+8x+3$ . Looking for a pair of factors of $12$ $12$ whose sum is $8$ $8$ , we see that $2, 6$ $2,6$ works:

$4 x^{2} + 8 x + 3$ $4x^2+8x+3$

$= 4 x^{2} + 2 x + 6 x + 3$ $=4x^2+2x+6x+3$

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

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

Thus,

$4 x^{3} - 13 x - 6 = (x - 2) (2 x + 3) (2 x + 1)$ $4x^3-13x-6=(x-2)(2x+3)(2x+1)$

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Is it possible to factor $y = 4 x^{3} - 13 x - 6$ $y=4x^3-13x-6$ ? If so, what are the factors?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

Is it possible to factor y=4x^3-13x-6 y=4x3−13x−6y=4x^3-13x-6 ? If so, what are the factors?

1 Answer

Explanation:

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Impact of this question

Is it possible to factor $y = 4 x^{3} - 13 x - 6$ $y=4x^3-13x-6$ ? If so, what are the factors?