How do you factor $y = 2 x^{2} + 11 x + 12$ $y=2x^2+11x+12$ ?

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Answer 1 · 2016-08-02T12:26:48

$y = (2 x + 3) (x + 4)$ $y = (2x + 3)(x+4)$

Find factors of 2 and 12 which add together to give 11.

find the cross products and add. There is some trial and error.

$2 3 \Rightarrow 1 \times 3 = 3$ $" 2 3" rArr 1xx3 = 3$
$1 4 \Rightarrow 2 \times 4 = 83 + 8 = 11$ $" 1 4" rArr 2xx4 = 8 " "3+8=11$

The signs are the same they are both positive.

$y = (2 x + 3) (x + 4)$ $y = (2x + 3)(x+4)$

Answer 2 · 2016-08-02T12:36:39

$(x + 4) (2 x + 3)$ $(x+4)(2x+3)$

First, let's find the zeros of the function y by solving the equation:

$2 x^{2} + 11 x + 12 = 0$ $2x^2+11x+12=0$

$x = \frac{- 11 \pm \sqrt{11^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$ $x=(-11+-sqrt(11^2-4*2*12))/(2*2)$

$x = \frac{- 11 \pm \sqrt{121 - 96}}{4}$ $x=(-11+-sqrt(121-96))/4$

$x = \frac{- 11 \pm \sqrt{25}}{4}$ $x=(-11+-sqrt(25))/4$

$x = \frac{- 11 \pm 5}{4}$ $x=(-11+-5)/4$

$x_{1} = - 4 and x_{2} = - \frac{3}{2}$ $x_1=-4 and x_2=-3/2$

You can factor a trynomial:

$y = a x^{2} + b x + c = a (x - x_{1}) (x - x_{2})$ $y=ax^2+bx+c=a(x-x_1)(x-x_2)$

and have:

$2 x^{2} + 11 x + 12 = 2 (x + 4) (x + \frac{3}{2})$ $2x^2+11x+12=2(x+4)(x+3/2)$

that's

$(x + 4) (2 x + 3)$ $(x+4)(2x+3)$

How do you factor y=2x2+11x+12y=2x^2+11x+12 ?