How do you simplify $\frac{4 y^{3} + 12 y^{2} - y - 3}{2 y^{3} + y^{2} - 18 y - 9}$ $(4y^3+12y^2-y-3)/(2y^3+y^2-18y-9)$ ?

Question

1999 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-04T01:52:14

$\frac{2 y - 1}{y - 3}$ $(2y-1)/(y-3)$

We can approach this by factoring:

$\frac{4 y^{3} + 12 y^{2} - y - 3}{2 y^{3} + y^{2} - 18 y - 9}$ $(4y^3+12y^2-y-3)/(2y^3+y^2-18y-9)$

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

$\frac{(4 y^{3} + 12 y^{2}) - (y + 3)}{(2 y^{3} + y^{2}) - (18 y + 9)}$ $((4y^3+12y^2)-(y+3))/((2y^3+y^2)-(18y+9))$

$\frac{4 y^{2} (y + 3) - (y + 3)}{y^{2} (2 y + 1) - 9 (2 y + 1)}$ $(4y^2(y+3)-(y+3))/(y^2(2y+1)-9(2y+1))$

$\frac{(4 y^{2} - 1) (y + 3)}{(y^{2} - 9) (2 y + 1)}$ $((4y^2-1)(y+3))/((y^2-9)(2y+1))$

We can now factor down the terms with $y^{2}$ $y^2$ :

$\frac{(2 y - 1) (2 y + 1) (y + 3)}{(y - 3) (y + 3) (2 y + 1)}$ $((2y-1)(2y+1)(y+3))/((y-3)(y+3)(2y+1))$

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)4y3+12y2−y−32y3+y2−18y−9(4y^3+12y^2-y-3)/(2y^3+y^2-18y-9)?