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Thread: factoring

  1. #1
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    factoring

    Hi all,
    why can't a^2+b^2 be factored in fact non of the following factor


    a^4+b^4
    a^8+b^8
    a^16+b^16

    the factors are all even numbers and some even factors can be broken up to to produce a prime number factor ie 10 become 2^5 but i can't quite see why these sums can't be factored.

    Thanks for the help.
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  2. #2
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    Remember that factoring allows you to solve equations: $\displaystyle a^2- b^2= 0$ can be solved by $\displaystyle (a- b)(a+ b)= 0$ so that a- b= 0 or a= b and a+ b= 0 or a= -b are solutions.

    If it were possible to factor $\displaystyle a^2+ b^2$ with real coefficients (we can factor it as $\displaystyle a^2+ b^2= a^2- (-b)^2= (a- bi)(a+ bi)$) then it would be possible to find non-zero a and b so that $\displaystyle a^2+ b^2= 0$ and of course, that is not true. Since the square of any non-zero number is positive, if a and b are not both 0, then $\displaystyle a^2+ b^2$ must be positive, not 0.
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  3. #3
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    thanks hallsofivy i also read something about there being no common factors can you help on this matter
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  4. #4
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    I have no idea what you mean by that. Are you still referring to $\displaystyle a^2+ b^2$?

    $\displaystyle a^2- b^2$ also has "no common factors" but can be factored.
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  5. #5
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    Why can a^2-b^2 be factored and a^2+b^2 can't I thought there needed to be a common factor(s) to factor something.
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  6. #6
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    No, you do not have to have "common factors"- at least not obvious ones.

    IF you have a "common factor", like the "a" in ab+ ac, then you can simply use the "distributive law": ab+ ac= a(b+ c). But with binomials or trinomials it can be a bit more complicated.

    While $\displaystyle a^2- b^2$ does NOT have an obvious "common factor", we could rewrite it as $\displaystyle a^2- ab+ ab- b^2= (a^2- ab)+ (ab- b^2$. Now the first binomial has a "common factor" of a and the second has a "common factor" of b: $\displaystyle a(a- b)+ b(a- b)$. And now we can see that the two terms have a "common factor" of a-b: $\displaystyle (a+ b)(a- b)$. But that is not always the simplest thing to do- for something like $\displaystyle a^2- b^2$ it is simplest to remember the simple formula, $\displaystyle a^2- b^2= (a- b)(a+ b)$.

    Notice that if we try to do the same thing with $\displaystyle a^2+ b^2$ we would have $\displaystyle a^2- b^2= a^2- ab+ ab+ b^2= a(a- b)+ b(a+ b)$ and now we cannot continue further.

    But the real reason we cannot factor $\displaystyle a^2+ b^2$ or, more generally, $\displaystyle a^{2n}+ b^{2n}$ is what I said before: $\displaystyle a^{2n}+ b^{2n}= 0$ cannot have any real roots except a= b= 0. If we were able to factor, with real coefficients, we could find an infinite number of real number values for a and b that would satisfy the equation.
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