Question

How do you simplify $ \dfrac{12{{x}^{7}}}{4{{x}^{2}}} $ ?

Answer 1

Hint: We have been given a division of functions of x. We first try to find the sign of the division from the coefficients. Then we divide the constants and the variables separately. For the division of the variables, we take the help of the identity of indices. We multiply them at the end to find the solution of $ \dfrac{12{{x}^{7}}}{4{{x}^{2}}} $ .

Complete step-by-step answer:
We have been given a division of two terms $ \dfrac{12{{x}^{7}}}{4{{x}^{2}}} $ .
The terms being $ 12{{x}^{7}} $ and $ 4{{x}^{2}} $ . Both of the terms are positive which makes the whole division positive.
Therefore, the sign is known to us. We just need to find the division of $ 12{{x}^{7}} $ and $ 4{{x}^{2}} $ .
We take the constants and variables separately. At the end we take multiplication of them.
Therefore, the form becomes $ \dfrac{12{{x}^{7}}}{4{{x}^{2}}}=\dfrac{12}{4}\times \dfrac{{{x}^{7}}}{{{x}^{2}}} $ .
We divide the constants first and get $ \dfrac{12}{4}=3 $ .
For example, we take two exponential expressions where the exponents are $ m $ and $ n $ .
Let the numbers be $ {{a}^{m}} $ and $ {{a}^{n}} $ . In case of division the indices get subtracted as $ \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} $ .
Then we divide the variables and get $ \dfrac{{{x}^{7}}}{{{x}^{2}}}={{x}^{7-2}}={{x}^{5}} $ .
Now we multiply them to get the final answer.
The total division becomes $ \dfrac{12{{x}^{7}}}{4{{x}^{2}}}=\dfrac{12}{4}\times \dfrac{{{x}^{7}}}{{{x}^{2}}}=3{{x}^{5}} $ . The sign is positive also.
The simplified form of $ \dfrac{12{{x}^{7}}}{4{{x}^{2}}} $ is $ 3{{x}^{5}} $ .
So, the correct answer is “ $ \dfrac{12{{x}^{7}}}{4{{x}^{2}}} $ is $ 3{{x}^{5}} $ ”.

Note: For any division and to find the appropriate signs for that we can follow the rule where division of same signs gives positive result and division of opposite signs give negative result. We can’t divide all the constants and the variables altogether to make the process simple. The division of a variable with constant is not possible.