# How do you divide (3a^3+17a^2+12a-5)/(a+5)?

Jul 14, 2017

$3 {a}^{2} + 2 a + 2 - \frac{15}{a + 5}$

#### Explanation:

First, let's split the top polynomial up into "multiples" of the bottom polynomial. To see what I mean, let's first try to take care of the $3 {a}^{3}$ term.

We can see that $3 {a}^{2} \left(a + 5\right) = 3 {a}^{3} + 15 {a}^{2}$. So, let's separate this from our polynomial:

$3 {a}^{3} + 17 {a}^{2} + 12 a - 5$

$\left(3 {a}^{3} + 15 {a}^{2}\right) + 2 {a}^{2} + 12 a - 5$

$3 {a}^{2} \left(a + 5\right) + 2 {a}^{2} + 12 a - 5$

See how this "gets rid of" the $3 {a}^{3}$ term? Let's do the same for the $2 {a}^{2}$ term.

We can see that $2 a \left(a + 5\right) = 2 {a}^{2} + 10 a$.

$3 {a}^{2} \left(a + 5\right) + \left(2 {a}^{2} + 10 a\right) + 2 a - 5$

$3 {a}^{2} \left(a + 5\right) + 2 a \left(a + 5\right) + 2 a - 5$

The next highest term to deal with is the $2 a$ term.

We can see that $2 \left(a + 5\right) = 2 a + 10$.

$3 {a}^{2} \left(a + 5\right) + 2 a \left(a + 5\right) + \left(2 a + 10\right) - 15$

$3 {a}^{2} \left(a + 5\right) + 2 a \left(a + 5\right) + 2 \left(a + 5\right) - 15$

We can't do anything about the $15$, since its degree is smaller than the degree of $a + 5$.

Finally, let's divide everything by $\left(a + 5\right)$. This is why we wrote everything in terms of $a + 5$, so we can just cancel out the top and the bottom of the fraction at this step.

$\frac{3 {a}^{3} + 17 {a}^{2} + 12 a - 5}{a + 5}$

$= \frac{3 {a}^{2} \left(a + 5\right) + 2 a \left(a + 5\right) + 2 \left(a + 5\right) - 15}{a + 5}$

$= 3 {a}^{2} + 2 a + 2 - \frac{15}{a + 5}$