# How do you find the quotient of (8k^2-6)div2k?

Oct 22, 2017

You would need to divide $2 k$ into each of the terms of the expression $\left(8 {k}^{2} - 6\right)$

$\frac{8 {k}^{2} - 6}{2 k} = 4 k - \frac{3}{k}$

#### Explanation:

$\frac{8 {k}^{2} - 6}{2 k} = \frac{8 {k}^{2}}{2 k} - \frac{6}{2 k} = 4 k - \frac{3}{k}$

Where we divided both integers by the $2$ and

subtracted a ${k}^{1}$ from both terms.

Since there was no $k$ in the second term, the subtraction of ${k}^{1}$ becomes a ${k}^{-} 1$ which is the same as $\frac{1}{k} ^ 1 \mathmr{and} \frac{1}{k}$

To check, randomly make $k = 3$:

$\frac{8 {k}^{2} - 6}{2 k} = 4 k - \frac{3}{k}$

$\frac{8 {\left(3\right)}^{2} - 6}{2 \left(3\right)} = 4 \left(3\right) - \frac{3}{3}$

$\frac{72 - 6}{6} = 12 - 1$

$11 = 11$