
How do you factor by grouping $4+9xy-18y-2x$?
Answer
548.1k+ views
Hint: To factorise the given polynomial by the method of factor by grouping, we first have to rearrange its terms such that on grouping the terms, we can take factors common from those groups of the terms. So we can rearrange the terms of the given polynomial and rewrite it as $-2x+4+9xy-18y$. Then we can combine the first and the last two terms together to make the two groups $\left( -2x+4 \right)+\left( 9xy-18y \right)$. From these groups, we can take the common factors of \[-2\] and \[9y\] outside to factorize it.
Complete step by step solution:
Let us consider the polynomial given in the above question as
$\Rightarrow p\left( x,y \right)=4+9xy-18y-2x$
For using the method of factor by grouping to factor the above polynomial, we first need to rearrange the terms of the above polynomial. So we rearrange the terms of the polynomial as
\[\Rightarrow p\left( x,y \right)=-2x+4+9xy-18y\]
Now, let us combine the first two and the last two terms so as to form two groups of the terms to get
\[\Rightarrow p\left( x,y \right)=\left( -2x+4 \right)+\left( 9xy-18y \right)\]
Now, the factors of \[-2\] and \[9y\] are common to the first and the second groups respectively. So we can take these outside of the respective groups to get
\[\Rightarrow p\left( x,y \right)=-2x\left( x-2 \right)+9y\left( x-2 \right)\]
Now, we can take the common factor \[\left( x-2 \right)\] outside to finally get
\[\begin{align}
& \Rightarrow p\left( x,y \right)=\left( x-2 \right)\left( -2x+9y \right) \\
& \Rightarrow p\left( x,y \right)=\left( x-2 \right)\left( 9y-2x \right) \\
\end{align}\]
Hence, we have factorized the given polynomial completely using the method of factor by grouping as \[\left( x-2 \right)\left( 9y-2x \right)\].
Note:
While rearranging the terms, make it sure that the first term of each of the two groups must be having a higher degree than the second term. For example, do not rearrange the terms as $4-2x+9xy-18y$ since in the first group, the degree of the first term $4$, equal to zero, is less than that of the first term $-2x$, which is equal to one.
Complete step by step solution:
Let us consider the polynomial given in the above question as
$\Rightarrow p\left( x,y \right)=4+9xy-18y-2x$
For using the method of factor by grouping to factor the above polynomial, we first need to rearrange the terms of the above polynomial. So we rearrange the terms of the polynomial as
\[\Rightarrow p\left( x,y \right)=-2x+4+9xy-18y\]
Now, let us combine the first two and the last two terms so as to form two groups of the terms to get
\[\Rightarrow p\left( x,y \right)=\left( -2x+4 \right)+\left( 9xy-18y \right)\]
Now, the factors of \[-2\] and \[9y\] are common to the first and the second groups respectively. So we can take these outside of the respective groups to get
\[\Rightarrow p\left( x,y \right)=-2x\left( x-2 \right)+9y\left( x-2 \right)\]
Now, we can take the common factor \[\left( x-2 \right)\] outside to finally get
\[\begin{align}
& \Rightarrow p\left( x,y \right)=\left( x-2 \right)\left( -2x+9y \right) \\
& \Rightarrow p\left( x,y \right)=\left( x-2 \right)\left( 9y-2x \right) \\
\end{align}\]
Hence, we have factorized the given polynomial completely using the method of factor by grouping as \[\left( x-2 \right)\left( 9y-2x \right)\].
Note:
While rearranging the terms, make it sure that the first term of each of the two groups must be having a higher degree than the second term. For example, do not rearrange the terms as $4-2x+9xy-18y$ since in the first group, the degree of the first term $4$, equal to zero, is less than that of the first term $-2x$, which is equal to one.
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