Question

How do you factor completely \[{{x}^{2}}-{{y}^{2}}+xz-yz\]?

Answer 1

Hint: From the given question we are asked to find the complete factorisation of the given expression \[{{x}^{2}}-{{y}^{2}}+xz-yz\]. For solving these kinds of questions we will simplify the terms in the expression by taking common terms outside as factors of the terms inside the brackets. We will also use the difference of squares identity and solve the given question completely.

Complete step by step answer:
Firstly, as we said above we will take the common factors. So, we get the expression simplified as follows.
\[\Rightarrow {{x}^{2}}-{{y}^{2}}+xz-yz\]
Now, we will take the \[z\] common in the expression for the last two that is third and fourth terms. So, we get the expression reduced as follows.
\[\Rightarrow {{x}^{2}}-{{y}^{2}}+z(x-y)\]
We know the identity in mathematics which is the difference of square identity.
The difference of squares identity will be as follows.
If we have two terms x and y then the identity will be,
\[\Rightarrow {{x}^{2}}-{{y}^{2}}=(x+y)(x-y)\]
So, we will use this identity and simplify the expression further as follows.
\[\Rightarrow {{x}^{2}}-{{y}^{2}}+z(x-y)\]
The first two terms in the expression can be simplified using the identity as follows.
\[\Rightarrow \left( x+y \right)\left( x-y \right)+z(x-y)\]
Now, in the above expression we can take the term \[\left( x-y \right)\] common from the two terms we are having in our simplified expression.
So, the expression will be further reduced as follows.
\[\Rightarrow \left( x+y+z \right)\left( x-y \right)\]

Therefore, the complete factorisation for the question is \[\Rightarrow \left( x+y+z \right)\left( x-y \right)\].

Note: Students must not make any calculation mistake. We must have good knowledge in basic algebra and must know the difference of squares identity which is \[\Rightarrow {{x}^{2}}-{{y}^{2}}=(x+y)(x-y)\] very well. We must not do mistake in doing the calculation like if we write solution as \[\Rightarrow \left( x-y+z \right)\left( x+y \right)\] instead of \[\Rightarrow \left( x+y+z \right)\left( x-y \right)\] then our solution will be wrong.