Factorize \[25{a^2} - 4{b^2} + 28bc - 49{c^2}\].
Answer
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Hint: In order to factorize the expression \[25{a^2} - 4{b^2} + 28bc - 49{c^2}\], we will try to manipulate the terms \[4{b^2} + 28bc - 49{c^2}\] to make it a perfect square using the algebraic identity \[{\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2}\]. Then we will use the algebraic identity \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\] to factorize the given expression.
Complete step by step answer:
We are given the expression,
\[25{a^2} - 4{b^2} + 28bc - 49{c^2}\]
We will try to rewrite the given expression in a way that it gets simplified.
On rewriting we get,
\[ = 25{a^2} - {\left( {2b} \right)^2} + 2 \times \left( {2b} \right) \times \left( {7c} \right) - {\left( {7c} \right)^2}\]
Using the algebraic identity \[{x^2} - 2xy + {y^2} = {\left( {x - y} \right)^2}\], we can write \[{\left( {2b} \right)^2} + 2 \times \left( {2b} \right) \times \left( {7c} \right) - {\left( {7c} \right)^2}\] as \[{\left( {2b - 7c} \right)^2}\].
Therefore, we can write the given expression as
\[ = 25{a^2} - {\left( {2b - 7c} \right)^2}\]
On rewriting the above expression, we get
\[ = {\left( {5a} \right)^2} - {\left( {2b - 7c} \right)^2}\]
As we know the algebraic identity \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\].
Here, \[x = 5a\] and \[y = 2b - 7c\].
Using this we can write the expression
\[ \Rightarrow {\left( {5a} \right)^2} - {\left( {2b - 7c} \right)^2} = \left[ {5a + \left( {2b + 7c} \right)} \right]\left[ {5a - \left( {2b + 7c} \right)} \right]\]
On simplifying, we get
\[ = \left( {5a + 2b + 7c} \right)\left( {5a - 2b - 7c} \right)\]
Therefore, factorization of \[25{a^2} - 4{b^2} + 28bc - 49{c^2}\] is \[\left( {5a + 2b + 7c} \right)\left( {5a - 2b - 7c} \right)\].
Note:
It is not necessary that there is only one method to factorize or only one method can be used at a time. Here, in this question we can see that we have first used the identity to simplify the expression then we have taken the common factors to factorize the given expression. It might be possible that various methods are involved in different steps to solve a question.
Complete step by step answer:
We are given the expression,
\[25{a^2} - 4{b^2} + 28bc - 49{c^2}\]
We will try to rewrite the given expression in a way that it gets simplified.
On rewriting we get,
\[ = 25{a^2} - {\left( {2b} \right)^2} + 2 \times \left( {2b} \right) \times \left( {7c} \right) - {\left( {7c} \right)^2}\]
Using the algebraic identity \[{x^2} - 2xy + {y^2} = {\left( {x - y} \right)^2}\], we can write \[{\left( {2b} \right)^2} + 2 \times \left( {2b} \right) \times \left( {7c} \right) - {\left( {7c} \right)^2}\] as \[{\left( {2b - 7c} \right)^2}\].
Therefore, we can write the given expression as
\[ = 25{a^2} - {\left( {2b - 7c} \right)^2}\]
On rewriting the above expression, we get
\[ = {\left( {5a} \right)^2} - {\left( {2b - 7c} \right)^2}\]
As we know the algebraic identity \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\].
Here, \[x = 5a\] and \[y = 2b - 7c\].
Using this we can write the expression
\[ \Rightarrow {\left( {5a} \right)^2} - {\left( {2b - 7c} \right)^2} = \left[ {5a + \left( {2b + 7c} \right)} \right]\left[ {5a - \left( {2b + 7c} \right)} \right]\]
On simplifying, we get
\[ = \left( {5a + 2b + 7c} \right)\left( {5a - 2b - 7c} \right)\]
Therefore, factorization of \[25{a^2} - 4{b^2} + 28bc - 49{c^2}\] is \[\left( {5a + 2b + 7c} \right)\left( {5a - 2b - 7c} \right)\].
Note:
It is not necessary that there is only one method to factorize or only one method can be used at a time. Here, in this question we can see that we have first used the identity to simplify the expression then we have taken the common factors to factorize the given expression. It might be possible that various methods are involved in different steps to solve a question.
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