Is it possible to factor $y = 3 x^{3} + 8 x^{2} + 3 x - 2$ $y= 3x^3+8x^2+3x-2$ ? If so, what are the factors?

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Is it possible to factor $y = 3 x^{3} + 8 x^{2} + 3 x - 2$ $y= 3x^3+8x^2+3x-2$ ? If so, what are the factors?

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Answer 1 · 2016-02-12T20:49:29

$3 x^{3} + 8 x^{2} + 3 x - 2 = (x + 1) (3 x - 1) (x + 2)$ $3x^3+8x^2+3x-2 = (x+1)(3x-1)(x+2)$

Explanation:

Let $f (x) = 3 x^{3} + 8 x^{2} + 3 x - 2$ $f(x) = 3x^3+8x^2+3x-2$

First notice that $f (- 1) = - 3 + 8 - 3 - 2 = 0$ $f(-1) = -3+8-3-2 = 0$ , so $x = - 1$ $x=-1$ is a zero and $(x + 1)$ $(x+1)$ is a factor:

$y = 3 x^{3} + 8 x^{2} + 3 x - 2 = (x + 1) (3 x^{2} + 5 x - 2)$ $y = 3x^3+8x^2+3x-2 = (x+1)(3x^2+5x-2)$

We can factor the remaining quadratic by grouping, with the help of an AC Method.

Look for a pair of factors of $A C = 3 \cdot 2 = 6$ $AC = 3*2 = 6$ which differ by $B = 5$ $B=5$ . The pair $6, 1$ $6, 1$ works, so use that to split the middle term before factoring by grouping:

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

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

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

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

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

Putting it all together:

$3 x^{3} + 8 x^{2} + 3 x - 2 = (x + 1) (3 x - 1) (x + 2)$ $3x^3+8x^2+3x-2 = (x+1)(3x-1)(x+2)$

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Is it possible to factor $y = 3 x^{3} + 8 x^{2} + 3 x - 2$ $y= 3x^3+8x^2+3x-2$ ? If so, what are the factors?

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Explanation:

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Is it possible to factor y=3x3+8x2+3x−2y= 3x^3+8x^2+3x-2 ? If so, what are the factors?

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Is it possible to factor $y = 3 x^{3} + 8 x^{2} + 3 x - 2$ $y= 3x^3+8x^2+3x-2$ ? If so, what are the factors?