# How do you divide (6x ^ { 3} + 17x ^ { 2} + 13x + 20) by ( 2x + 5)?

Oct 11, 2017

Quotient $3 {x}^{2} + x + 4$, Remainder $0$

#### Explanation:

Denominator may be written as 2*(x+(5/2) so that, first we can divide by (x+(5/2)) and then by 2 as a second step.
……………………………………………………………………
-(5/2)…….|...…6……17…….13………20
……………..|………….-15……-5……..-20
……………..|_________
…………………..6…….2…….8……….0

We have divided by $\left(x + \left(\frac{5}{2}\right)\right)$ where as we have to divide it by $\left(2 x + 5\right)$ which is nothing but $2 \cdot \left(x + \left(\frac{5}{2}\right)\right)$
So we should divide the quotient by 2 to get the answer.
$\frac{6 {x}^{2} + 2 x + 8}{2} = 3 {x}^{2} + x + 4$ which is the actual quotient and Remainder remains 0.

Quotient $\frac{6 {x}^{2} + 2 x + 8}{2}$, Remainder $0$

Oct 11, 2017

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

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

Division is part and parcel of multiplication.

This approach is really the same as LinaM but looks different.

The problem is that marking schemes for exams are very strict and (particularly in UK) are based on a particular approach. If that approach is not used then there is every chance you will not get method marks. Method marks are what you are after. The correct answer is the icing on the cake. If you have the wrong answer but are awarded method marks then your attempt is not a total loss.

The approach to use is that shown to you by your teacher.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering your question}}$

$\textcolor{w h i t e}{\text{ddddddddddddd}} 6 {x}^{3} + 17 {x}^{2} + 13 x + 20$
$\textcolor{m a \ge n t a}{3 {x}^{2}} \left(2 x + 5\right) \textcolor{w h i t e}{\text{d")->ul(6x^3+15x^2 larr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddd}} 0 + 2 {x}^{2} + 13 x + 20$
$\textcolor{m a \ge n t a}{x} \left(2 x + 5\right) \textcolor{w h i t e}{\text{d")->color(white)("ddddddd")ul(2x^2+5xlarr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddddddd.d}} 0 + 8 x + 20$
$\textcolor{m a \ge n t a}{4} \left(2 x + 5\right) \textcolor{w h i t e}{\text{d")->color(white)("dddd.dddddd.d")ul( 8x+20 larr" Subtract}}$
No remainder so stop$\textcolor{w h i t e}{\text{ddddddddd}} 0 + 0$

$\textcolor{m a \ge n t a}{3 {x}^{2} + x + 4}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$

$\textcolor{w h i t e}{\text{dddddd}} 3 {x}^{2} + x + 4$
color(white)("ddddd")ul(color(white)("ddddd")2x+5)
$\textcolor{w h i t e}{d} 6 {x}^{3} + 2 {x}^{2} + 8 x$
color(white)("d")ul(color(white)("dddd")15x^2+5x+20color(white)("ddd")) larr" Add"#
$6 {x}^{3} + 17 {x}^{2} + 13 x + 20$ as required.

color(white)("d")