MathsLearner
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- Aug 13, 2017
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Very difficult question do not know how to proceed. Please provide me some hints to solve it.
Resolve \(\displaystyle x^8\, +\, y^8\) into real quadratic factors.
What have you tried? Where are you stuck? Please show us some work so we can guide you. What makes you think that this is even factorable???Very difficult question do not know how to proceed. Please provide me some hints to solve it.
Resolve \(\displaystyle x^8\, +\, y^8\) into real quadratic factors.
But I don't know the answer for this? How should i know it is factorable? Please help.What makes you think that this is even factorable???
This is gonna sound kinda stupid and flippant but, while you generally would not probably look at this and immediately think, "Gee, sure, yes; this is factorable!", the fact that they told you to factor it generally means that (while a pain in the hindquarters) the polynomial must be factorable....What makes you think that this is even factorable???
But I don't know the answer for this? How should i know it is factorable?
Well, the Fundamental Theorem of Algebra says that is at least factorable into quadratics with real coefficients because it is a polynomial of even degree with real coefficients. Of course, that theorem says nothing about how easy it is to find such a factoring.What have you tried? Where are you stuck? Please show us some work so we can guide you. What makes you think that this is even factorable???
Very difficult question do not know how to proceed. Please provide me some hints to solve it.
Resolve \(\displaystyle x^8\, +\, y^8\) into real quadratic factors.
A factoring is a product, and your answer is not a product.Thank you for the hints. I have just did it like attached.
\(\displaystyle \bigg(\left(x^4\right)\, -\, \left(y^4\right)\bigg)^2\, =\, \left(x^4\right)^2\, +\, \left(y^4\right)^2\, -\, 2\, x^4\, y^4\)
\(\displaystyle x^8\, +\, y^8\, =\, \bigg(\left(x^2\, +\, y^2\right)\, \left(x^2\, -\, y^2\right)\bigg)^2\, +\, 2\, \left(x^2\, y^2\right)^2\)