Question

Factorise the given expression : ${x^2} - xz + xy - yz$.

Answer 1

Hint: Observe the given expression carefully. Find out the common factors if any and take it out. Then we get the factors of the given expression. Here we can take $x$ common from the first two terms and $y$ common from the last two terms. Then we again get a common factor. Taking that too common outside we can factorise the given expression to two factors.

Complete step-by-step answer:
Consider the given expression: ${x^2} - xz + xy - yz$.
We can see that $x$ is common in the first two terms and $y$ is common in the last two terms.
Taking them common outside from the terms we get,
${x^2} - xz + xy - yz = x(x - z) + y(x - z)$
Now we can see that there is again a common factor which is $x - z$.
So we can take it outside from both the terms.
Thus we get,
${x^2} - xz + xy - yz = (x - z)(x + y)$
Therefore the given expression is factored to two linear factors $x - z$ and $x + y$.

Additional information:
If while solving we get expressions like ${a^2} - {b^2}$ we can factorise it.
For we have,
${a^2} - {b^2} = (a + b)(a - b)$
Also we have the results,
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${(a - b)^2} = {a^2} - 2ab + {b^2}$
Similarly we have the formulas including cube terms and so on.


Note: We can solve the problem in other ways as well.
That is, we get the same factors with a different approach.
Consider the given expression ${x^2} - xz + xy - yz$.
With slight rearrangement of the terms we get, ${x^2} + xy - xz - yz$
Now we can take $x$ common from the first two terms and $z$ common from the last two terms.
Thus we get, ${x^2} + xy - xz - yz = x(x + y) - z(x + y)$.
So here we have the common term $x + y$.
Taking it common outside we get,
${x^2} + xy - xz - yz = (x + y)(x - z)$
This gives,
${x^2} - xz + xy - yz = (x + y)(x - z)$
Therefore we get the same factors here.