Question

Factorise the following expressions:
(1) \[6a+6b\]
(2) \[12{{x}^{3}}y-4x{{y}^{2}}\]
(3) \[3ay+3az\]

Answer 1

Hint: We solve this problem by factorizing the given expressions one by one.
We take the common terms out from each term of the expression and then we check whether further factorization is possible or not.
If it is possible then we use the required algebra formula to factorize the given expressions.

Complete step by step answer:
We are given some expressions to factorize.
Let us factorize the given expressions one by one.
(1) \[6a+6b\]
Let us assume that the given expression as
\[\Rightarrow E=6a+6b\]
Now, let us take the common terms out from the above equation then we get
\[\Rightarrow E=6\left( a+b \right)\]
Here, we can see that the above expression cannot be further factorised
So, we can conclude that the factorisation of \[6a+6b\] as
\[\therefore 6a+6b=6\left( a+b \right)\]
(2) \[12{{x}^{3}}y-4x{{y}^{2}}\]
Let us assume that the given expression as
\[\Rightarrow A=12{{x}^{3}}y-4x{{y}^{2}}\]
Now, let us take the common terms out from the above equation then we get
\[\Rightarrow A=4xy\left( 3{{x}^{2}}-y \right)\]
Here, we can see that the above expression cannot be further factorised
So, we can conclude that the factorisation of \[12{{x}^{3}}y-4x{{y}^{2}}\] as
\[\therefore 12{{x}^{3}}y-4x{{y}^{2}}=4xy\left( 3{{x}^{2}}-y \right)\]
(3) \[3ay+3az\]
Let us assume that the given expression as
\[\Rightarrow P=3ay+3az\]
Now, let us take the common terms out from the above equation then we get
\[\Rightarrow P=3a\left( y+z \right)\]
Here, we can see that the above expression cannot be further factorized
So, we can conclude that the factorization of \[3ay+3az\] as
\[\therefore 3ay+3az=3a\left( y+z \right)\]

Note:
Students may do mistakes in the factorization method.
In the first expression, we get the expression after the factorization as
\[\Rightarrow E=6\left( a+b \right)\]
Here, we can stop the factorization because there is no possibility of further factorization.
But students may do mistake and take the expression further as
\[\Rightarrow E=6\left( {{\left( \sqrt{a} \right)}^{2}}-{{\left( \sqrt{b} \right)}^{2}} \right)\]
Now, we can factorize further. But this is no need because we mostly try to factorize the given expression in the linear form that is having the power of variable as 1. So, there is no need for this type of further factorization.