Question

How do you factor $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy$?

Answer 1

Hint: We first explain the process of factorisation which is available for $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy$. We find the variables which can be taken as common from the individual terms. We take $xy$ common and form the equation as the multiplication of terms.

Complete step by step answer:
The given equation $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy$ is a polynomial of two 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 $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy$.
Now we are actually taking $xy$ common from the terms of $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy$. We are taking $xy$ as common from the terms $3{{x}^{3}}{{y}^{2}},-2{{x}^{2}}y,5xy$.
If we take $xy$ from $3{{x}^{3}}{{y}^{2}}$, the quotient will be $\dfrac{3{{x}^{3}}{{y}^{2}}}{xy}=3{{x}^{2}}y$.
If we take $xy$ from $-2{{x}^{2}}y$, the quotient will be $\dfrac{-2{{x}^{2}}y}{xy}=-2x$.
If we take $xy$ from $5xy$, the quotient will be $\dfrac{5xy}{xy}=5$.
Now the common term $xy$ will form the multiplication of two terms. One being $xy$ and the other being the addition of $3{{x}^{2}}y,-2x,5$.
The addition gives $3{{x}^{2}}y-2x+5$

Therefore, the factorisation is $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy=xy\left( 3{{x}^{2}}y-2x+5 \right)$.

Note: We can verify the result of the factorisation by taking an arbitrary value of $x,y$ where $x=1,y=2$.
We put $x=1,y=2$ on the left-hand side of the equation $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy=xy\left( 3{{x}^{2}}y-2x+5 \right)$ and get
$\begin{align}
  & 3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy \\
 & =3\times {{1}^{3}}\times {{2}^{2}}-2\times {{1}^{2}}\times 2+5\times 1\times 2 \\
 & =12-4+10 \\
 & =18 \\
\end{align}$
Now we put $x=1,y=2$ on the right-hand side of the equation $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy=xy\left( 3{{x}^{2}}y-2x+5 \right)$ and get
$\begin{align}
  & xy\left( 3{{x}^{2}}y-2x+5 \right) \\
 & =1\times 2\left( 3\times {{1}^{2}}\times 2-2\times 1+5 \right) \\
 & =2\times 9 \\
 & =18 \\
\end{align}$.
Thus, verified the factorisation $3{{x}^{3}}{{y}^{2}}-2{{x}^{2}}y+5xy=xy\left( 3{{x}^{2}}y-2x+5 \right)$.