How do you find all the zeros of #f(x) = 9x^2 + 6x - 8#?

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Answer 1 · 2016-08-05T22:43:47

The zeros of #f(x)# are #x=2/3# and #x=-4/3#

#f(x) = 9x^2+6x-8#

Use an AC method to factor #f(x)#:

Find a pair of factors of #AC=9*8=72# with difference #B=6#.

The pair #12, 6# works.

Use this pair to split the middle term and factor by grouping:

#9x^2+6x-8#

#=9x^2+12x-6x-8#

#=(9x^2+12x)-(6x+8)#

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

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

Hence the zeros of #f(x)# are #x=2/3# and #x=-4/3#