How do you factor $y = x^{3} + 8 x^{2} + 19 x + 12$ $y=x^3 + 8x^2 + 19x + 12$ ?

Question

How do you factor $y = x^{3} + 8 x^{2} + 19 x + 12$ $y=x^3 + 8x^2 + 19x + 12$ ?

1 Answer

Impact of this question

1495 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-15T21:00:21

$(x + 1) (x + 4) (x - 3)$ $(x+1)(x+4)(x-3)$

Explanation:

First, use all integer negative numbers that divide the known term 12 and substitute them in x
Prove x=-1
$y = {(- 1)}^{3} + 8 \cdot {(- 1)}^{2} + 19 \cdot (- 1) + 12 = 0$ $y=(-1)^3+8*(-1)^2+19*(-1)+12=0$
then use Ruffini method to have:

$(x^{2} + 7 x + 12) (x + 1)$ $(x^2+7x+12)(x+1)$

$x = \frac{- 7 \pm \sqrt{49 - 48}}{2}$ $x=(-7+-sqrt(49-48))/2$
$x = \frac{- 7 \pm 1}{2}$ $x=(-7+-1)/2$
$x = - 4 and x = 3$ $x=-4 and x=3$

Science

Math

Humanities

... and beyond

How do you factor $y = x^{3} + 8 x^{2} + 19 x + 12$ $y=x^3 + 8x^2 + 19x + 12$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor y=x3+8x2+19x+12y=x^3 + 8x^2 + 19x + 12 ?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor $y = x^{3} + 8 x^{2} + 19 x + 12$ $y=x^3 + 8x^2 + 19x + 12$ ?