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

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Answer 1 · 2016-08-02T12:26:48

#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" rArr 1xx3 = 3#
#" 1 4" rArr 2xx4 = 8 " "3+8=11#

The signs are the same they are both positive.

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

Answer 2 · 2016-08-02T12:36:39

#(x+4)(2x+3)#

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

#2x^2+11x+12=0#

#x=(-11+-sqrt(11^2-4*2*12))/(2*2)#

#x=(-11+-sqrt(121-96))/4#

#x=(-11+-sqrt(25))/4#

#x=(-11+-5)/4#

#x_1=-4 and x_2=-3/2#

You can factor a trynomial:

#y=ax^2+bx+c=a(x-x_1)(x-x_2)#

and have:

#2x^2+11x+12=2(x+4)(x+3/2)#

that's

#(x+4)(2x+3)#