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anemone
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Factorize $x^2+y^2+z^2-2xy-2yz-2zx$.
Very Well done, lfdahl! (Cool)lfdahl said:My attempt:
In order to perform the factorization, I must require: $y,z \ge 0$.
\[x^2+y^2+z^2-2xy-2zx-2yz \\\\= (x-y-z)^2-4yz \\\\=(x-y-z+2\sqrt{yz})(x-y-z-2\sqrt{yz}) \\\\=(x-((\sqrt{y})^2+(\sqrt{z})^2-2\sqrt{yz}))(x-((\sqrt{y})^2+(\sqrt{z})^2+2\sqrt{yz})) \\\\=(x-(\sqrt{y}-\sqrt{z})^2)(x-(\sqrt{y}+\sqrt{z})^2)\]
The "Factorization Challenge" is a mathematical problem-solving activity that involves breaking down a number into its prime factors.
Factorization is the process of finding all the numbers that can evenly divide into a given number. These numbers are called factors, and when multiplied together, they equal the original number.
Factorization is an essential concept in mathematics and is used in many fields such as cryptography, number theory, and algebra. It allows us to simplify complex expressions and solve equations.
The goal of the "Factorization Challenge" is to find the prime factors of a given number in the shortest amount of time and with the fewest number of steps. It is a fun and challenging way to practice and improve your factorization skills.
The best way to improve your factorization skills is to practice regularly. You can also use various techniques such as trial division, prime factorization using a factor tree, or using a calculator. Additionally, participating in activities like the "Factorization Challenge" can also help improve your skills.