Factoring the Sum of Two Fifth Powers

In summary, the conversation discusses different attempts at factoring the expression x^5 + y^5, with the correct answer being (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4). The conversation also mentions using the sum of cubes formula and its limitations in this case.
  • #1
magisbladius
7
0
My problem is to factor [tex]x^{5}[/tex] + [tex]y^{5}[/tex].

I get [tex](x+y)(x^{8}-x^{4}y^{4}+y^{8})[/tex].

However, the answer is [tex](x+y)(x^{4}-x^{2}y^{2}+y^{4})[/tex].
 
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  • #2
Neither is right! Multiplying out in either cases does not give you the original back.

The correct answer is (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
 
  • #3
[tex]x^{5}+y^{5}[/tex]
[tex]=x^{3}x^{2}+y^{3}y^{2}[/tex]​

[tex]=(x^{2}x+y^{2}y)(x^{4}x^{2}-x^{2}xy^{2}y+y^{4}y^{2})[/tex]​

[tex]=(x+y)(x^{2}-xy+y^{2})(x^{6}-x^{3}y^{3}+y^{6})[/tex]​

[tex]=(x+y)(x^{8}-x^{5}y^{3}+x^{2}y^{6}-x^{7}y+x^{4}y^{3}-xy^{7}+x^{6}y^{2}-x^{3}y^{5}+y^{8})[/tex]​

What am I doing wrong?

------------------------

Your answer:
[tex]=(x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})[/tex]​
 
  • #4
magisbladius said:
What am I doing wrong?
The first and last steps are the only ones that make any sense; I have no idea what you were trying to do inbetween.
 
  • #5
How about you try factoring 5*x^4*y + 10*x^3*y^2 + 10* x^2 * y^3 + 5*x*y^4 instead? This gives 5*xy( (x+y)^3 - x^2*y - x*y^2) = 5*xy( (x+y)^3 - xy(x + y)) = 5*xy*(x+y)*((x+y)^2 - xy). Since x^5 + y^5 = (x+y)^5 - (5*x^4*y + 10*x^3*y^2 + 10* x^2 * y^3 + 5*x*y^4), we obtain x^5 + y^5 = (x+y)^5 - 5*xy*(x+y)*((x+y)^2 - xy) = (x+y)((x+y)^4 - 5*xy*(x+y)*((x+y)^2 - xy)), and take it from there (you can extract (x+y) still).
 
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  • #6
You could think of that expression as [/itex]x^5 - (-y)^5[/itex] and apply that into [tex]a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-1})[/tex]. You can verify that general equation by expanding the RHS.
 
  • #7
magisbladius said:
[tex]x^{5}+y^{5}[/tex]
[tex]=x^{3}x^{2}+y^{3}y^{2}[/tex]​

[tex]=(x^{2}x+y^{2}y)(x^{4}x^{2}-x^{2}xy^{2}y+y^{4}y^{2})[/tex]​


What am I doing wrong?

------------------------

Your answer:
[tex]=(x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})[/tex]​

Your last assertion makes no sense. You have terms x9 and y9 among others. They seem to come out of nowhere!
 
  • #8
mathman said:
Your last assertion makes no sense. You have terms x9 and y9 among others. They seem to come out of nowhere!

I thought applying the sum of cubes on the cubed power parts would have worked [tex]x^{3}x^{2}+y^{3}y^{2}[/tex].

Now I realize why that was inappropriate.
 
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1. What is factoring the sum of two fifth powers?

Factoring the sum of two fifth powers is a mathematical process that involves finding the factors of an expression that is written as the sum of two terms, each raised to the fifth power. This can also be referred to as factoring a fifth degree polynomial.

2. Why is factoring the sum of two fifth powers important?

Factoring the sum of two fifth powers is important because it allows us to simplify complex expressions and solve equations more easily. It also helps us understand the relationships between different mathematical concepts and can be used to find patterns in numbers.

3. How do I factor the sum of two fifth powers?

To factor the sum of two fifth powers, we can use the difference of squares formula (a^2 - b^2 = (a+b)(a-b)) and the sum of cubes formula (a^3 + b^3 = (a+b)(a^2-ab+b^2)). We can also use long division or synthetic division to factor higher degree polynomials.

4. Can the sum of two fifth powers be factored into two terms?

Yes, the sum of two fifth powers can be factored into two terms. This is because the fifth power is an odd number, so when it is raised to an even exponent (such as 2), it becomes positive and when it is raised to an odd exponent (such as 3), it retains its negative sign. Therefore, the sum of two fifth powers will always have a common factor that can be factored out.

5. Are there any real life applications of factoring the sum of two fifth powers?

Yes, factoring the sum of two fifth powers has many real life applications. It is used in cryptography to encrypt and decrypt messages, in engineering to solve complex equations, and in economics to analyze financial data. It is also used in physics to model and understand natural phenomena.

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