# How do you simplify (6t^3 + 5t^2 + 9) / (2t + 3)?

May 10, 2018

$3 {t}^{2} - 2 {t}^{2} + 3$

#### Explanation:

$\frac{6 {t}^{3} + 5 {t}^{2} + 9}{2 t + 3}$

$\frac{2 t + 3 \left(3 {t}^{2} - 2 t + 3\right)}{2 t + 3}$

• 2t+3 on numerator and denomoniator cancel out.

ANSWER: $3 {t}^{2} - 2 {t}^{2} + 3$

You have to factorise the top that $2 t + 3$ directly factorises $6 {t}^{3} + 5 {t}^{2} + 9$ so that it cancels the denominator giving you a value of $3 {t}^{2} - 2 {t}^{2} + 3$.

The are many ways of factorising $6 {t}^{3} + 5 {t}^{2} + 9$, polynomial division, rational root theorem.

I would recommend polynomial division as its pretty straight forward to learn and rely on. Here is a link that will teach you all about it and you can practise with: http://www.purplemath.com/modules/polydiv2.htm

May 10, 2018

$3 {t}^{2} - 2 t + 3$

#### Explanation:

$\text{one way is to use the divisor as a factor in the numerator}$

$\text{consider the numerator}$

$\textcolor{red}{3 {t}^{2}} \left(2 t + 3\right) \textcolor{m a \ge n t a}{- 9 {t}^{2}} + 5 {t}^{2} + 9$

$= \textcolor{red}{3 {t}^{2}} \left(2 t + 3\right) \textcolor{red}{- 2 t} \left(2 t + 3\right) \textcolor{m a \ge n t a}{+ 6 t} + 9$

$= \textcolor{red}{3 {t}^{2}} \left(2 t + 3\right) \textcolor{red}{- 2 t} \left(2 t + 3\right) \textcolor{red}{+ 3} \left(2 t + 3\right) \cancel{\textcolor{m a \ge n t a}{- 9}} \cancel{+ 9}$

$= \textcolor{red}{3 {t}^{2}} \left(2 t + 3\right) \textcolor{red}{- 2 t} \left(2 t + 3\right) \textcolor{red}{+ 3} \left(2 t + 3\right) + 0$

$\Rightarrow \frac{\cancel{\left(2 t + 3\right)} \left(3 {t}^{2} - 2 t + 3\right)}{\cancel{\left(2 t + 3\right)}} = 3 {t}^{2} - 2 t + 3$