How do you simplify #(4y^3+12y^2-y-3)/(2y^3+y^2-18y-9)#?

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Answer 1 · 2017-02-04T01:52:14

#(2y-1)/(y-3)#

We can approach this by factoring:

#(4y^3+12y^2-y-3)/(2y^3+y^2-18y-9)#

Looking at the first two terms in both the numerator and denominator, they appear to be similar to the last two terms:

#((4y^3+12y^2)-(y+3))/((2y^3+y^2)-(18y+9))#

#(4y^2(y+3)-(y+3))/(y^2(2y+1)-9(2y+1))#

#((4y^2-1)(y+3))/((y^2-9)(2y+1))#

We can now factor down the terms with #y^2#:

#((2y-1)(2y+1)(y+3))/((y-3)(y+3)(2y+1))#

And now we can cancel like terms:

#((2y-1)cancel((2y+1))cancel((y+3)))/((y-3)cancel((y+3))cancel((2y+1)))#

#(2y-1)/(y-3)#