How do you factor the trinomial #2x^2+ 11x-9#?

1 Answer
George C.
Jun 1, 2017

#2x^2+11x-9 = 1/8(4x+11-sqrt(193))(4x+11+sqrt(193))#

Explanation:

Given:

#2x^2+11x-9#

First note that this would factor really quickly if the constant was #+9# instead of #-9#:

#2x^2+11x+9 = 2x^2+2x+9x+9#

#color(white)(2x^2+11x+9) = 2x(x+1)+9(x+1)#

#color(white)(2x^2+11x+9) = (2x+9)(x+1)#

It is still possible to factor it using irrational coefficients.

Here's one way by completing the square and using the difference of squares identity:

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

#8(2x^2+11x-9) = 16x^2+88x-72#

#color(white)(8(2x^2+11x-9)) = (4x)^2+2(4x)(11)+121-193#

#color(white)(8(2x^2+11x-9)) = (4x+11)^2-(sqrt(193))^2#

#color(white)(8(2x^2+11x-9)) = ((4x+11)-sqrt(193))((4x+11)+sqrt(193))#

#color(white)(8(2x^2+11x-9)) = (4x+11-sqrt(193))(4x+11+sqrt(193))#

So:

#2x^2+11x-9 = 1/8(4x+11-sqrt(193))(4x+11+sqrt(193))#

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