# How do you divide (2x^3+7x^2-5x-4) / (2x+1) using polynomial long division?

Jun 15, 2018

Quotient is ${x}^{2} + 3 x - 4$ and remainder is $0$

#### Explanation:

$2 {x}^{3} + 7 {x}^{2} - 5 x - 4$

=$2 {x}^{3} + {x}^{2} + 6 {x}^{2} + 3 x - 8 x - 4$

=${x}^{2} \cdot \left(2 x + 1\right) + 3 x \cdot \left(2 x + 1\right) - 4 \cdot \left(2 x + 1\right)$

=$\left(2 x + 1\right) \cdot \left({x}^{2} + 3 x - 4\right)$

Hence quotient is ${x}^{2} + 3 x - 4$ and remainder is $0$

Jun 15, 2018

${x}^{2} + 3 x - 4$

#### Explanation:

Numerator $\to \textcolor{w h i t e}{\text{d}} 2 {x}^{3} + 7 {x}^{2} - 5 x - 4$
color(magenta)(x^2)(2x+1)-> color(white)("d")ul(2x^3+x^2larr" Subtract"
$\textcolor{w h i t e}{\text{dddddddddddddd}} 0 + 6 {x}^{2} - 5 x - 4$
$\textcolor{m a \ge n t a}{3 x} \left(2 x + 1\right) \to \textcolor{w h i t e}{\text{dddddd") ul(6x^2+3xlarr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddddddd}} 0 - 8 x - 4$
$\textcolor{m a \ge n t a}{- 4} \left(2 x + 1\right) \to \textcolor{w h i t e}{\text{ddddddd.d")ul(-8x-4 larr" Subtract}}$
"Remainder"-> color(white)("ddddddddddd")0color(white)("d")+0

$\textcolor{w h i t e}{\text{dddddddddddddd}} \textcolor{m a \ge n t a}{{x}^{2} + 3 x - 4}$