How do you solve #2x^3 + 6x^2 – x – 3 = 0#?

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Answer 1 · 2016-01-12T20:47:19

Factor by grouping and using the difference of squares identity to find:

#x=+-sqrt(2)/2# or #x=-3#

The difference of squares identity can be written:

#a^2-b^2=(a-b)(a+b)#

We use this below with #a=sqrt(2)x# and #b=1#.

Factor by grouping then use difference of squares identity:

#0 = 2x^3+6x^2-x-3#

#=(2x^3+6x^2)-(x+3)#

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

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

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

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

So #x=+-1/sqrt(2) = +-sqrt(2)/2# or #x=-3#

graph{ 2x^3+6x^2-x-3 [-10.84, 9.16, -3.60, 7.2]}