Question

How do you factor $2pqrs+14{{q}^{2}}rs-49qrs$?

Answer 1

Hint: We first explain the process of factorisation which is available for $2pqrs+14{{q}^{2}}rs-49qrs$. We find the variables which can be taken as common from the individual terms. We take $qrs$ common and form the equation as the multiplication of terms.

Complete step by step solution:
The given equation $2pqrs+14{{q}^{2}}rs-49qrs$ is a polynomial of three variables.
We need to factorise the equation.
The only process that is available for this equation to factorise is to take a common out of the terms from $2pqrs+14{{q}^{2}}rs-49qrs$.
Now we are actually taking $qrs$ common from the terms of $2pqrs+14{{q}^{2}}rs-49qrs$. We are taking $qrs$ as common from the terms $2pqrs,14{{q}^{2}}rs,49qrs$.
If we take $qrs$ from $2pqrs$, the quotient will be $\dfrac{2pqrs}{qrs}=2p$.
If we take $qrs$ from $14{{q}^{2}}rs$, the quotient will be $\dfrac{14{{q}^{2}}rs}{qrs}=14q$.
If we take $qrs$ from $-49qrs$, the quotient will be $\dfrac{-49qrs}{qrs}=-49$.
Now the common term $qrs$ will form the multiplication of two terms. One being $qrs$ and the other being the addition of $2p,14q,-49$.
Therefore, the factorisation is $2pqrs+14{{q}^{2}}rs-49qrs=qrs\left( 2p+14q-49 \right)$.

Note: We can verify the result of the factorisation by taking an arbitrary value of $p,q,r,s$ where $p=1,q=2,r=3,s=4$.
We put $p=1,q=2,r=3$ on the left-hand side of the equation $2pqrs+14{{q}^{2}}rs-49qrs$ and get
$\begin{align}
  & 2pqrs+14{{q}^{2}}rs-49qrs \\
 & =2\times 1\times 2\times 3\times 4+14\times {{2}^{2}}\times 3\times 4-49\times 2\times 3\times 4 \\
 & =-456 \\
\end{align}$
Now we put $p=1,q=2,r=3$ on the right-hand side of the equation $qrs\left( 2p+14q-49 \right)$ and get
$\begin{align}
  & qrs\left( 2p+14q-49 \right) \\
 & =2\times 3\times 4\left( 2\times 1+14\times 2-49 \right) \\
 & =-456 \\
\end{align}$.
Thus, verified the factorisation $2pqrs+14{{q}^{2}}rs-49qrs=qrs\left( 2p+14q-49 \right)$.