Question

Write the quotient and remainder of the following division: $\left( -48{{p}^{4}} \right)\div \left( -9{{p}^{2}} \right)$?

Answer 1

Hint: We start solving the problem by finding the dividend and divisor in the division. We then recall the standard form of a dividend. We then write the dividend as a sum of two terms. We then factorize the terms and convert the given dividend into the standard form. We get the values of quotient and remainder after converting into the standard form of a dividend.

Complete step-by-step solution
According to the problem, we need to find the quotient and remainder from the division of $\left( -48{{p}^{4}} \right)\div \left( -9{{p}^{2}} \right)$.
From the problem, we can see that the dividend is $-48{{p}^{4}}$ and the divisor is $-9{{p}^{2}}$. Let us convert the dividend into the standard from involving divisor, quotient and remainder i.e., $\text{dividend= }\left( \text{divisor}\times \text{quotient} \right)\text{ + remainder}$ ---(1).
Let us write the dividend in terms of the sum of two terms.
i.e., $-48{{p}^{4}}=-45{{p}^{4}}-3{{p}^{4}}$.
Let us factorize the term $45{{p}^{4}}$. We know that 45 can be written as $\left( 9\times 5 \right)=\left( 3\times 3\times 5 \right)$. Now, we can write the terms ${{p}^{4}}$ can be written as $p\times p\times p\times p$. Since, p is a variable we cannot factorize further as the absolute value of p is unknown.
So, the factorization of $45{{p}^{4}}$ is $\left( 3\times 3\times 5\times p\times p\times p\times p \right)$
$\Rightarrow -48{{p}^{4}}=\left( -3\times 3\times 5\times p\times p\times p\times p \right)-3{{p}^{4}}$. (Here – is multiplied in order to satisfy the negative sign present in the term).
$\Rightarrow -48{{p}^{4}}=\left( \left( -3\times 3\times p\times p \right)\times \left( 5\times p\times p \right) \right)-3{{p}^{4}}$.
$\Rightarrow -48{{p}^{4}}=\left( -9{{p}^{2}} \right)\times \left( 5{{p}^{2}} \right)-3{{p}^{4}}$.
$\Rightarrow -48{{p}^{4}}=\left( -9{{p}^{2}} \right)\times \left( 5{{p}^{2}} \right)+\left( -3{{p}^{4}} \right)$ ---(2).
We can see that equation (2) resembles equation (1) as the dividend is $-48{{p}^{4}}$ and the divisor is $-9{{p}^{2}}$. Now, we find quotient and remainder by comparing both equations (1) and (2).
So, we get the quotient as $5{{p}^{2}}$ and remainder as $-3{{p}^{2}}$.
$\therefore$The quotient and remainder of the division $\left( -48{{p}^{4}} \right)\div \left( -9{{p}^{2}} \right)$ is $5{{p}^{2}}$ and $-3{{p}^{2}}$.

Note: We should not say quotient as $\dfrac{16}{3}{{p}^{2}}$ and remainder 0, as it is the fraction representation of the given division process. We can also solve this by using the long division method as shown below:
$\begin{align}
  & \left. -9{{p}^{2}} \right)-48{{p}^{4}}\left( 5{{p}^{2}} \right. \\
 & \text{ }-45{{p}^{2}} \\
 & \underline{\text{ }\left( + \right)\text{ }} \\
 & \text{ }-3{{p}^{2}} \\
\end{align}$.
Since we don't know the absolute value of p, we should go by the standard process. If we know the absolute value of p, we should make sure that the value of the remainder must be positive.