How do you simplify \[\dfrac{\dfrac{4{{t}^{2}}-16}{8}}{\dfrac{t-2}{6}}\] ?

Hint: These types of problems are easy solved by taking terms common and cancelling them out. We start by cancelling the arithmetic terms first, and then proceed to the algebraic terms. In this problem, as the expression is quite large, we divide it into two parts. Having divided it into two parts, we carry on the steps on each of the parts separately. Finally, we recombine the two evaluated terms to get the final term. In this problem, the key role is played by taking terms common and cancelling them.

Complete step by step answer:
The above expression is
\[\dfrac{\dfrac{4{{t}^{2}}-16}{8}}{\dfrac{t-2}{6}}\]
Let us assume,
\[x=\dfrac{a}{b}\]
Where, \[a=\dfrac{4{{t}^{2}}-16}{8}\]and \[b=\dfrac{t-2}{6}\] .
We first need to simplify \[a\]. Let us take 4 common in the numerator. Then,
\[a=\dfrac{4\left( {{t}^{2}}-4 \right)}{8}\]
Dividing numerator and denominator by 4, we get
\[a=\dfrac{\left( {{t}^{2}}-4 \right)}{2}\]
Factorising \[\left( {{t}^{2}}-4 \right)\] we get
\[a=\dfrac{\left( t+2 \right)\left( t-2 \right)}{2}\]
Now putting the above values in the original equation , \[x\] becomes
\[x=\dfrac{\dfrac{\left( t+2 \right)\left( t-2 \right)}{2}}{\dfrac{\left( t-2 \right)}{6}}\]
Multiplying with 6 in the numerator and denominator we get,
\[x=\dfrac{3\left( t+2 \right)\left( t-2 \right)}{\left( t-2 \right)}\]
Cancelling out the \[\left( t-2 \right)\]terms from the numerator and denominator, we get
\[x=3\left( t+2 \right)\]

Thus, the given expression is simplified to \[3\left( t+2 \right)\]

Note: Fractions always involve performing errors while evaluating them. Therefore, we got to be careful while solving the fractions. The most common mistake that students make is that they get confused after seeing such a huge expression and overlap one’s numerator with others’. This common mistake can easily be avoided by separating the terms out at first, solving them separately, and recombining them at the end. We should also be careful while cancelling out the terms and should always prefer to cancel out the arithmetic terms first.