Question

Factorize the following: \[{x^2} + {y^2} + {z^2} - xy - yz - zx\]

Answer 1

Hint: We will multiply and divide the polynomial by a constant so that we can simplify it. We will try to change them in any algebraic formulas. We should be familiar with formulas like ${(a - b)^2} = {a^2} + {b^2} - 2ab$ , \[{a^2} - {b^2} = (a + b)(a - b)\] etc.

Complete answer:
We have given \[{x^2} + {y^2} + {z^2} - xy - yz - zx\]
We will multiply and divide the polynomial by 2, because it will not affect the polynomial
\[ = \dfrac{2}{2}\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)\]
We will take 2inside the bracket and multiply the polynomial
\[ = \dfrac{1}{2}\left( {2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx} \right)\]
We will expand each term such that we can make them in the form of perfect squares.
\[ = \dfrac{1}{2}\left( {{x^2} + {x^2} + {y^2} + {y^2} + {z^2} + {z^2} - 2xy - 2yz - 2zx} \right)\]
We know that ${(a - b)^2} = {a^2} + {b^2} - 2ab$
\[ = \dfrac{1}{2}\left[ {\left( {{x^2} + {y^2} - 2xy} \right) + \left( {{y^2} + {z^2} - 2yz} \right) + \left( {{x^2} + {z^2} - 2zx} \right)} \right]\]
\[ = \dfrac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {x - z} \right)}^2}} \right]\]
Hence, the factorization of \[{x^2} + {y^2} + {z^2} - xy - yz - zx\] is \[\dfrac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {x - z} \right)}^2}} \right]\]

Note:
We have to be very careful in solving these types of questions because most of the questions can only be solved by only one method easily. It may be solved by other methods too but it will become very lengthy. And the best approach will come with a lot of practice and by remembering formulas.