How do you factor #2x^2+9x-5?

1 Answer
Alan P.
May 5, 2015

#2x^2+9x-5#

Hoping for an integer factoring, we look for factors of #2# and factors of #5# which when cross multiplied then subtracted give #9#
(we know we need to subtract since the final term (#5#) is negative).

Fortunately there are not a lot of possible options for the factors and we soon come up with

#{: (,2,xx,5,=,10),(,1,xx,-1,=-,1),(,,,,,9):}#

So the factors of #2x^2+9x-5#
are #(2x-1)(x+5)#

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