How do you factor #y=x^2 + 8x - 9# ?

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Answer 1 · 2016-04-02T13:54:01

#y=x^2+8x-9" "->" "y=(x+9)(x-1)#

As there is no coefficient of #x^2# so we are only looking for factors of 9 where their difference is 8.

Consider the factors 1 and 9

We know that #1xx9=9#. So that is ok!

We also know that #9-1=8# So that is ok as well!

So we have the general form of #(x+?)(x+?)#

The 8 from #8x# is positive so for the difference we need 9 to be positive and the 1 to be negative so that we end up with #9-1=8#

#color(blue)((x+9)color(brown)((x-1))#

Just as a check lets multiply out the brackets

#color(brown)(color(blue)(x)(x-1)color(blue)(+9)(x-1))#

#x^2-x+9x-9#

#x^2+8x-9color(red)(" Works ok!")#
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#y=x^2+8x-9" "->" "y=(x+9)(x-1)#