Question

How do you factor by grouping $ab+2a+3b+6$?

Answer 1

Hint: The polynomial given in the above question, which is $ab+2a+3b+6$, consists of four terms, which means it consists of even number of terms. So we can use the factor by grouping method to combine two-two terms so as to form two groups of the terms. In the given polynomial, we can combine the terms as $\left( ab+2a \right)+\left( 3b+6 \right)$, so that the first group is $\left( ab+2a \right)$ and the second group is $\left( 3b+6 \right)$. The factors $a$ and $3$ can be taken from these respective groups. After that, we will have one factor common to both the terms. On taking that outside, the given polynomial will be factored.

Complete step by step solution:
The polynomial given in the above question is $ab+2a+3b+6$. It is in terms of $a$ and $b$. Therefore, we can consider it as
$\Rightarrow p\left( a,b \right)=ab+2a+3b+6$
Now, since the polynomial consists of even number of terms, we can combine two-two terms to form two groups as
$\Rightarrow p\left( a,b \right)=\left( ab+2a \right)+\left( 3b+6 \right)$
Now, we can take $a$ and $3$ common from the two groups in the above polynomial to get
$\Rightarrow p\left( a,b \right)=a\left( b+2 \right)+3\left( b+2 \right)$
Finally, we take the common factor $\left( b+2 \right)$ outside of the above polynomial to get
\[\Rightarrow p\left( a,b \right)=\left( b+2 \right)\left( a+3 \right)\]
Hence, we have used the factor by grouping method to factor the given polynomial as \[\left( b+2 \right)\left( a+3 \right)\].

Note:
After taking the terms common from all of the groups of the terms formed in the beginning, it is important that we get a factor common to all the groups. If we are not getting any factor common to all the groups, then either we need to reconsider the combination of the terms to form the groups, or we have to check for the mistakes in taking terms common.