Question

Divide the following: $-21ab{{c}^{2}}$ by $7abc$.

Answer 1

Hint: We start solving the problem by factorizing the given number $-21ab{{c}^{2}}$. After factorizing $-21ab{{c}^{2}}$, we make arrangements in the factors and made it look like the division form $\text{dividend=}\left( \text{divisor}\times \text{quotient} \right)+\text{remainder}$. On comparing dividend and divisor, we can see that those are clearly coinciding with the values that are given in the problem. This makes us get the required values of quotient and remainder.

Complete step-by-step solution:
According to the problem, we need to divide $-21ab{{c}^{2}}$ with $7abc$.
Let us first factorize $-21ab{{c}^{2}}$ in order to make division.
We know that the number 21 can be written as $3\times 7$. As the numbers 3 and 7 are prime, we cannot factorize 21 further. So, we get $21=3\times 7$ --------(1).
We know that $ab{{c}^{2}}$ can be written as $a\times b\times c\times c$. As a, b and c are variables and the absolute values of them are unknown, we cannot factorize them further. So, we get $ab{{c}^{2}}=a\times b\times c\times c$ --------(2).
From equations (1) and (2), we get factorization of $21ab{{c}^{2}}$ as $3\times 7\times a\times b\times c\times c$. But we need the factorization of $-21ab{{c}^{2}}$ so, we multiply the $-1$ to the obtained factorization result of $21ab{{c}^{2}}$.
So, we get factorization of $-21ab{{c}^{2}}$ as $-1\times 3\times 7\times a\times b\times c\times c$.
Now, we have $-21ab{{c}^{2}}=-1\times 3\times 7\times a\times b\times c\times c$.
$\Rightarrow -21ab{{c}^{2}}=\left( 7\times a\times b\times c \right)\times \left( -1\times 3\times c \right)$.
$\Rightarrow -21ab{{c}^{2}}=\left( 7abc \right)\times \left( -3c \right)$.
$\Rightarrow -21ab{{c}^{2}}=\left( \left( 7abc \right)\times \left( -3c \right) \right)+0$ --------(3).
We can see that the equation (3) resembles the division process $\text{dividend=}\left( \text{divisor}\times \text{quotient} \right)+\text{remainder}$. Comparing both we get:
$\Rightarrow $ Dividend = $-21ab{{c}^{2}}$.
$\Rightarrow $ Divisor = $7abc$.
$\Rightarrow $ Quotient = $-3c$.
$\Rightarrow $ Remainder = 0.
According to the problem, we have to divide $-21ab{{c}^{2}}$ by $7abc$ which clearly makes $-21ab{{c}^{2}}$ as dividend and $7abc$ as divisor.
So, we have found quotient and remainder as $-3c$ and 0.
$\therefore$ The quotient and remainder obtained on dividing $-21ab{{c}^{2}}$ with $7abc$ is $-3c$ and 0.

Note: We can see that this is one of the ways to solve the problem. We can also solve this problem by first using the law of exponents taking $-21ab{{c}^{2}}$ in the numerator and $7abc$ in the denominator. We can then use the factorization results to get the required value of the quotient. We know that if the denominator divides the numerator and gives the answer without any decimal in it, then the remainder is 0.