# How do you divide (3x^2+7x-20) / (x+4)?

Jul 30, 2015

$3 x - 5$

#### Explanation:

Hint: If you see the question carefully, you might notice that the question was intended so that you may have to factor the numerator in such a way that $x + 4$ is a factor, and gets cancelled with the denominator. Let's try doing that.

Let $N = 3 {x}^{2} + 7 x - 20$.
Note that $3 x \left(x + 4\right) = 3 {x}^{2} + 12 x$

Thus, we write:
$N = 3 {x}^{2} + 12 x - 5 x - 20 = 3 x \left(x + 4\right) - 5 \left(x + 4\right) = \left(3 x - 5\right) \left(x + 4\right)$

Now we come to the question itself
$\frac{3 {x}^{2} + 7 x - 20}{x + 4} = \frac{\left(3 x - 5\right) \cancel{\left(x + 4\right)}}{\cancel{\left(x + 4\right)}} = 3 x - 5$