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

1 Answer
Sep 23, 2016

Answer:

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

Explanation:

Note that if the #-# sign was a #+#, then this would factor with integer coefficients:

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

To factor the given quadratic with a minus sign, we can complete the square, but to cut down on fractions, I would like to multiply through by #8# first, then divide through by #8# at the end...

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

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