
How do you factor \[2x + 2y + ax + ay\]?
Answer
493.8k+ views
Hint: The process of writing an expression as a product of two or more expressions is called factorization and those individual expressions are called as factors of the given expression. When we divide a polynomial by its factor the remainder is zero. To find the factor of the given expression we will first group the terms and then take common and express the expression as a product of two expressions.
Complete step-by-step solution:
The given expression is:
\[2x + 2y + ax + ay\]
Now we will group the terms. So, we get;
\[ = \left( {2x + 2y} \right) + \left( {ax + ay} \right)\]
On observing carefully, we can see that \[2\] is common in the first bracket and \[a\] is common in the second bracket. So, we will take the common terms out of the bracket.
\[ = 2\left( {x + y} \right) + a\left( {x + y} \right)\]
Now we can see that \[\left( {x + y} \right)\] is common in both the terms. So, we will take it as common and we get,
\[ = \left( {x + y} \right)\left( {2 + a} \right)\]
\[\therefore 2x + 2y + ax + ay = \left( {x + y} \right)\left( {2 + a} \right)\]
Hence, \[\left( {x + y} \right),\left( {2 + a} \right)\] are the factors of \[2x + 2y + ax + ay\].
Additional Information:
One can also solve this question by grouping the terms differently. For example, we can also make the group as:
\[2x + 2y + ax + ay = \left( {2x + ax} \right) + \left( {2y + ay} \right)\]
On taking common from each group, we get;
\[ = x\left( {2 + a} \right) + y\left( {2 + a} \right)\]
On further taking \[\left( {2 + a} \right)\] as common we get;
\[ = \left( {2 + a} \right)\left( {x + y} \right)\]
Similarly, other groups can also be formed but the answer will be the same.
Note: One important thing should note here is that when we deal with the numbers, the factors are always less than or equal to the given number and when we deal with polynomials the degree of factors are less than or equal to the degree of the given polynomial. One interesting thing to note is that the greatest factor of a number is the number itself.
Complete step-by-step solution:
The given expression is:
\[2x + 2y + ax + ay\]
Now we will group the terms. So, we get;
\[ = \left( {2x + 2y} \right) + \left( {ax + ay} \right)\]
On observing carefully, we can see that \[2\] is common in the first bracket and \[a\] is common in the second bracket. So, we will take the common terms out of the bracket.
\[ = 2\left( {x + y} \right) + a\left( {x + y} \right)\]
Now we can see that \[\left( {x + y} \right)\] is common in both the terms. So, we will take it as common and we get,
\[ = \left( {x + y} \right)\left( {2 + a} \right)\]
\[\therefore 2x + 2y + ax + ay = \left( {x + y} \right)\left( {2 + a} \right)\]
Hence, \[\left( {x + y} \right),\left( {2 + a} \right)\] are the factors of \[2x + 2y + ax + ay\].
Additional Information:
One can also solve this question by grouping the terms differently. For example, we can also make the group as:
\[2x + 2y + ax + ay = \left( {2x + ax} \right) + \left( {2y + ay} \right)\]
On taking common from each group, we get;
\[ = x\left( {2 + a} \right) + y\left( {2 + a} \right)\]
On further taking \[\left( {2 + a} \right)\] as common we get;
\[ = \left( {2 + a} \right)\left( {x + y} \right)\]
Similarly, other groups can also be formed but the answer will be the same.
Note: One important thing should note here is that when we deal with the numbers, the factors are always less than or equal to the given number and when we deal with polynomials the degree of factors are less than or equal to the degree of the given polynomial. One interesting thing to note is that the greatest factor of a number is the number itself.
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