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:

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

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

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