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

$\frac{2 y - 1}{y - 3}$

#### Explanation:

We can approach this by factoring:

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

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

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

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

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

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

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

And now we can cancel like terms:

$\frac{\left(2 y - 1\right) \cancel{\left(2 y + 1\right)} \cancel{\left(y + 3\right)}}{\left(y - 3\right) \cancel{\left(y + 3\right)} \cancel{\left(2 y + 1\right)}}$

$\frac{2 y - 1}{y - 3}$