
Factorize the expression: \[{{a}^{2}}-{{b}^{2}}+2bc-{{c}^{2}}\]
Answer
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Hint: When we factorize an algebraic expression, we write it as a product of factors. These factors may be number algebraic, variables, or algebraic expressions. An expression like \[4xy,3{{x}^{2}}y\]is already in the factored form. Their factors can be just read off from them, as we already know. On the other hand, consider expressions like \[4x+3,5x-2\]. It is not obvious what their factors are.
Complete step-by-step solution:
Factorization by regrouping terms: When the expression will not be easy to see the factorization. Rearranging the expression allows us to form groups to factorizations. This is regrouping.
We have given the equation \[{{a}^{2}}-{{b}^{2}}+2bc-{{c}^{2}}\]. The given expression is not in factor form, firstly we have to do rearranging the terms. In the given expression, there are four terms.
Check if there is a common factor among all terms. There is none.
Think of grouping: Notice the first two terms, in the first term \[{{a}^{2}}\] we can write as \[a\times a\]. In the second term \[{{b}^{2}}\], we can think \[b\times b\]. So, notice that in the first two terms they don’t have common factors.
So, keeping \[{{a}^{2}}\]term aside.
Notice that second term, third term and fourth term, \[-{{b}^{2}}+2bc-{{c}^{2}}\]
Taking minus symbol as common from above, we get as
\[-{{b}^{2}}+2bc-{{c}^{2}}=-\left( {{b}^{2}}-2bc+{{c}^{2}} \right)\]…………… (1)
Here, let us remember one basic formula which is \[{{\left( a-b \right)}^{2}}=\left( {{a}^{2}}-2ab+{{b}^{2}} \right)\]
Equation 1 can be written as \[-{{\left( b-c \right)}^{2}}\]
Combining the whole, we get
\[\Rightarrow {{a}^{2}}-{{\left( b-c \right)}^{2}}\]…………………. (2)
Again, we need to remember another basic formula, \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
equation (2) can be written as,
\[\Rightarrow \left( a+b-c \right)\left( a-\left( b-c \right) \right)\]
On evaluating, we get
\[\Rightarrow \left( a+b-c \right)\left( a-b+c \right)\]
The required answer is \[\left( a+b-c \right)\left( a-b+c \right)\].
Note: When finding the factors of a polynomial expression always try to find the constants and variables that are common in all the terms of that expression then take out those common elements and then write them outside as factors of that polynomial.
Complete step-by-step solution:
Factorization by regrouping terms: When the expression will not be easy to see the factorization. Rearranging the expression allows us to form groups to factorizations. This is regrouping.
We have given the equation \[{{a}^{2}}-{{b}^{2}}+2bc-{{c}^{2}}\]. The given expression is not in factor form, firstly we have to do rearranging the terms. In the given expression, there are four terms.
Check if there is a common factor among all terms. There is none.
Think of grouping: Notice the first two terms, in the first term \[{{a}^{2}}\] we can write as \[a\times a\]. In the second term \[{{b}^{2}}\], we can think \[b\times b\]. So, notice that in the first two terms they don’t have common factors.
So, keeping \[{{a}^{2}}\]term aside.
Notice that second term, third term and fourth term, \[-{{b}^{2}}+2bc-{{c}^{2}}\]
Taking minus symbol as common from above, we get as
\[-{{b}^{2}}+2bc-{{c}^{2}}=-\left( {{b}^{2}}-2bc+{{c}^{2}} \right)\]…………… (1)
Here, let us remember one basic formula which is \[{{\left( a-b \right)}^{2}}=\left( {{a}^{2}}-2ab+{{b}^{2}} \right)\]
Equation 1 can be written as \[-{{\left( b-c \right)}^{2}}\]
Combining the whole, we get
\[\Rightarrow {{a}^{2}}-{{\left( b-c \right)}^{2}}\]…………………. (2)
Again, we need to remember another basic formula, \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
equation (2) can be written as,
\[\Rightarrow \left( a+b-c \right)\left( a-\left( b-c \right) \right)\]
On evaluating, we get
\[\Rightarrow \left( a+b-c \right)\left( a-b+c \right)\]
The required answer is \[\left( a+b-c \right)\left( a-b+c \right)\].
Note: When finding the factors of a polynomial expression always try to find the constants and variables that are common in all the terms of that expression then take out those common elements and then write them outside as factors of that polynomial.
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