How do you find the quotient of (8c^3+6c-5)div(4c-2) using long division?

Jun 6, 2017

The remainder is $= - 1$ and the quotient is $= 2 {c}^{2} + c + 2$

Explanation:

Let's perform the long division

$4 c - 2$$\textcolor{w h i t e}{a a a a}$$|$$8 {c}^{3} + 0 {c}^{2} + 6 c - 5$$\textcolor{w h i t e}{a a a a}$$|$$2 {c}^{2} + c + 2$

$\textcolor{w h i t e}{a a a a a a a a a a a}$$8 {c}^{3} - 4 {c}^{2}$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$0 + 4 {c}^{2} + 6 c$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$+ 4 {c}^{2} - 2 c$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$+ 0 + 8 c - 5$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$+ 8 c - 4$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a}$$+ 0 - 1$

Therefore,

$\frac{8 {c}^{3} + 0 {c}^{2} + 6 c - 5}{4 c - 2} = 2 {c}^{2} + c + 2 - \frac{1}{4 c - 2}$

Jun 6, 2017

$\left(8 {c}^{3} + 6 c - 5\right) \div \left(4 c - 2\right) = 2 {c}^{2} + c + 2 \text{ rem } 1$

or:

$\left(8 {c}^{3} + 6 c - 5\right) \div \left(4 c - 2\right) = 2 {c}^{2} + c + 2 + \frac{- 1}{4 c - 2}$

Explanation:

• In each case, $4 c$ is divided into the term with the highest power of $c$ available.
• the quotient is multiplied by both terms at the side.
• change signs to subtract the expressions. (shown in red)

$\textcolor{w h i t e}{w w w w w w w w} 2 {c}^{2} + c + 2$
4c-2 )bar(8c^3 " "+6c-5)larr" " desc order, no term in ${c}^{2}$
$\textcolor{w h i t e}{w w w} \underline{\textcolor{red}{-} 8 {c}^{3} \textcolor{red}{+} 4 {c}^{2}} \text{ } \leftarrow$ subtract
$\textcolor{w h i t e}{w w w w w w w w w} 4 {c}^{2} + 6 c$
$\textcolor{w h i t e}{w w n \ldots . . w w} \textcolor{red}{-} \underline{4 {c}^{2} \textcolor{red}{+} 2 c}$
$\textcolor{w h i t e}{w w w w w w w w w w w} + 8 c - 5$
$\textcolor{w h i t e}{w w w w w w w w w w w} \textcolor{red}{-} \underline{8 c \textcolor{red}{+} 4}$
$\textcolor{w h i t e}{w w w w w w w w w w w w w w} - 1 \text{ }$ remainder

$\left(8 {c}^{3} + 6 c - 5\right) \div \left(4 c - 2\right) = 2 {c}^{2} + c + 2 \text{ rem } 1$

or:

$\left(8 {c}^{3} + 6 c - 5\right) \div \left(4 c - 2\right) = 2 {c}^{2} + c + 2 + \frac{- 1}{4 c - 2}$