How do you factor #2x^2+3x-2#?

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Answer 1 · 2015-05-22T15:04:44

You can find its roots and then turn them into factors. I"ll use Bhaskara to find the roots.

#(-3+-sqrt(9-4(2)(-2)))/4#
#(-3+-5)/4#
#x_1=-2#, which is the same as the factor #x+2=0#
#x_2=1/2#, which is the same as the factor #2x-1=0#

Thus,

#2x^2+3x-2=(x+2)(2x-1)#

Answer 2 · 2015-05-29T09:29:50

#2x^2 + 3x - 2#

Product = - 4
Sum = +3
Factors = 4, - 1
=

So spiting it up

#2x^2 - x + 4x - 2#

#x ( 2x - 1)+2(2x - 1)#
#(x + 2)(2x -1)#
=

Hence factored