How do you factor #(3x^2+4x-15)/(2x^2+3x-9)#?

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Answer 1 · 2017-06-16T20:45:22

#(3x-5)/(2x-3)#

We can break the top polynomial like this:

#3x^2 + 4x - 15#

#3x^2 + 9x - 5x - 15#

And then we can simplify it like this:

#(3x^2 + 9x) - (5x+15)#

#3x(x+3) - 5(x+3)#

#= (3x-5)(x+3)#

We can also break the bottom polynomial similarly like this:

#2x^2 + 3x - 9#

#2x^2 + 6x - 3x - 9#

#(2x^2 + 6x) - (3x+9)#

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

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

Therefore, our factored expression becomes:

#(3x^2 + 4x - 15)/(2x^2 + 3x - 9) = ((3x-5)(x+3))/((2x-3)(x+3)) = ((3x-5)cancel((x+3)))/((2x-3)cancel((x+3))#

After cancelling out the top and bottom of the expression, we get the result:

#(3x-5)/(2x-3)#

Final Answer