# Very difficult question: Resolve x^8 + y^8 into real quadratic factors.

#### MathsLearner

##### New member
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.

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#### Subhotosh Khan

##### Super Moderator
Staff member
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.
x^8 + y^8

=x^8 + y^8 + 2 * x^4 * y^4 - 2 * x^4 * y^4

continue....

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#### Jomo

##### Elite Member
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???

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#### MathsLearner

##### New member
Thank you for the hints. I have just did it like attached.

$$\displaystyle \bigg(\left(x^4\right)\, -\, \left(y^4\right)\bigg)^2\, =\, \left(x^4\right)^2\, +\, \left(y^4\right)^2\, -\, 2\, x^4\, y^4$$

$$\displaystyle x^8\, +\, y^8\, =\, \bigg(\left(x^2\, +\, y^2\right)\, \left(x^2\, -\, y^2\right)\bigg)^2\, +\, 2\, \left(x^2\, y^2\right)^2$$

What makes you think that this is even factorable???

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#### stapel

##### Super Moderator
Staff member
...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?
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.

#### JeffM

##### Elite Member
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???
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.

#### Dr.Peterson

##### Elite Member
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.
Let's back up a bit. This is not an easy problem; and the wording of it suggests that you have learned about the fact that any polynomial with real coefficients can be factored into linear or quadratic factors. Perhaps you have also been taught some other ideas that can help. What facts have you recently learned that might be relevant -- theorems, techniques, examples, ...? In order to help you do something, we really need to know what tools are in your toolbox!

#### JeffM

##### Elite Member
Thank you for the hints. I have just did it like attached.

$$\displaystyle \bigg(\left(x^4\right)\, -\, \left(y^4\right)\bigg)^2\, =\, \left(x^4\right)^2\, +\, \left(y^4\right)^2\, -\, 2\, x^4\, y^4$$

$$\displaystyle x^8\, +\, y^8\, =\, \bigg(\left(x^2\, +\, y^2\right)\, \left(x^2\, -\, y^2\right)\bigg)^2\, +\, 2\, \left(x^2\, y^2\right)^2$$
A factoring is a product, and your answer is not a product.

One way to do a factoring (which does not always help) is to add a needed term and subtract it. In this case, let's try to create a difference of two squares.

$$\displaystyle x^8 + y^ 8 = x^8 + y^8 + 0 = x^8 + y^8 + (2x^4y^4 - 2x^4y^4) =$$

$$\displaystyle (x^8 + 2x^4y^4 + y^8) - 2x^4y^4 = (x^4 + y^4)^2 - (\sqrt{2x^4y^4})^2=$$

$$\displaystyle (x^4 + y^4)^2 - (x^2y^2\sqrt{2})^2.$$

Once we have a difference of two squares, we can factor that.

$$\displaystyle x^8 + y^ 8 = (x^4 + y^4)^2 - (x^2y^2\sqrt{2})^2 = (x^4 + y^4 - x^2y^2\sqrt{2})(x^4 + y^4 + x^2y^2\sqrt{2}).$$

Now you have factored into two quartics. Each can be factored into two quadratics using the same technique. The algebra gets messy so make sure to check your answer.

LATE EDIT:

The quartics can also be factored using a substitution of variables and completion of the square.

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