
How do you factor by grouping $ {m^2} - {n^2} + 5m - 5n $ ?
Answer
537.3k+ views
Hint: To solve this question, we will group the first two terms and the second two terms. These both groups can be factorized in different ways. We can solve the first group by using the method of difference of Squares and other groups by taking common terms.
Formula used:
$ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $
Complete step-by-step answer:
We are given $ {m^2} - {n^2} + 5m - 5n $ .
Let us first divide the polynomial in two groups by keeping them in two different brackets.
$ \Rightarrow \left( {{m^2} - {n^2}} \right) + \left( {5m - 5n} \right) $
Now let us first consider the first group $ \left( {{m^2} - {n^2}} \right) $ . We can see that this is the difference of two squares. Therefore, to factorize it, we can use the formula $ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $ .
$ \Rightarrow \left( {{m^2} - {n^2}} \right) = \left( {m - n} \right)\left( {m + n} \right) $
Now, we will consider the second group $ \left( {5m - 5n} \right) $ .
We can see that 5 is the common digit in both the terms so that we can take it out as common.
Therefore, we can write Combining these both we can write the polynomial as:
$
{m^2} - {n^2} + 5m - 5n \\
= \left( {{m^2} - {n^2}} \right) + \left( {5m - 5n} \right) \\
= \left( {m - n} \right)\left( {m + n} \right) + 5\left( {m - n} \right) \;
$
We can see that the term $ \left( {m - n} \right) $ is common so that we can take it as common.
$
\Rightarrow {m^2} - {n^2} + 5m - 5n \\
= \left( {{m^2} - {n^2}} \right) + \left( {5m - 5n} \right) \\
= \left( {m - n} \right)\left( {m + n} \right) + 5\left( {m - n} \right) \\
= \left( {m - n} \right)\left( {m + n + 5} \right) \;
$
Thus our final answer is: $ \left( {m - n} \right)\left( {m + n + 5} \right) $ .
So, the correct answer is “ $ \left( {m - n} \right)\left( {m + n + 5} \right) $ ”.
Note: In this type of question, there are three general steps to be followed: First step is to group the first two terms together and then the last two terms together. The second step is to factor out a greatest common factor from each separate binomial which is $ \left( {m - n} \right) $ in our case. Finally, the third step is to factor out the common binomial as we have done in our last step which is to take out the term $ \left( {m - n} \right) $ as it is common.
Formula used:
$ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $
Complete step-by-step answer:
We are given $ {m^2} - {n^2} + 5m - 5n $ .
Let us first divide the polynomial in two groups by keeping them in two different brackets.
$ \Rightarrow \left( {{m^2} - {n^2}} \right) + \left( {5m - 5n} \right) $
Now let us first consider the first group $ \left( {{m^2} - {n^2}} \right) $ . We can see that this is the difference of two squares. Therefore, to factorize it, we can use the formula $ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $ .
$ \Rightarrow \left( {{m^2} - {n^2}} \right) = \left( {m - n} \right)\left( {m + n} \right) $
Now, we will consider the second group $ \left( {5m - 5n} \right) $ .
We can see that 5 is the common digit in both the terms so that we can take it out as common.
Therefore, we can write Combining these both we can write the polynomial as:
$
{m^2} - {n^2} + 5m - 5n \\
= \left( {{m^2} - {n^2}} \right) + \left( {5m - 5n} \right) \\
= \left( {m - n} \right)\left( {m + n} \right) + 5\left( {m - n} \right) \;
$
We can see that the term $ \left( {m - n} \right) $ is common so that we can take it as common.
$
\Rightarrow {m^2} - {n^2} + 5m - 5n \\
= \left( {{m^2} - {n^2}} \right) + \left( {5m - 5n} \right) \\
= \left( {m - n} \right)\left( {m + n} \right) + 5\left( {m - n} \right) \\
= \left( {m - n} \right)\left( {m + n + 5} \right) \;
$
Thus our final answer is: $ \left( {m - n} \right)\left( {m + n + 5} \right) $ .
So, the correct answer is “ $ \left( {m - n} \right)\left( {m + n + 5} \right) $ ”.
Note: In this type of question, there are three general steps to be followed: First step is to group the first two terms together and then the last two terms together. The second step is to factor out a greatest common factor from each separate binomial which is $ \left( {m - n} \right) $ in our case. Finally, the third step is to factor out the common binomial as we have done in our last step which is to take out the term $ \left( {m - n} \right) $ as it is common.
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