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Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$ Here is what I did:

$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$ $$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$ $$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$ $$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$ $$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$

At this point I don't know what to do and am feeling that my direction is wrong. Please help me.
( Wolfram alpha says that the answer is $(a^2-a b+b^2) (2 a^2+2 a b+b^2)$ but how? )

A Googler
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4 Answers4

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Different Hint

Since your polynomial is homogeneous, this is equivalent to factoring the degree 4 univariate polynomial $2x^4+x^2+x+1.$

minar
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Hint

We look for a factorization on the form

$$2a^4 + a^2b^2 + ab^3 + b^4=(2a^2+\alpha ab+b^2)(a^2+\beta ab+b^2)$$

We find $\alpha=2,\beta=-1$.

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    Looks tricky ;-) – Mikasa Aug 11 '13 at 17:02
  • From this hint it was easy to find the values of alpha and beta but my question is how did you realize that the factorization will be like that ? What makes you think that there will be two quadratic factors and not four linear or one linear and three cubic factors ? Would you be able to figure this out if you didn't know this factorization ? If yes , how? – A Googler Aug 11 '13 at 17:06
  • @AGoogler You have just to try all the possibilities (which aren't fortunately too much) and with some practice you'll find the form immediately. –  Aug 11 '13 at 17:11
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    Using minar's hint, you can quickly determine (rational root theorem) that there are no linear factors. – The Chaz 2.0 Aug 11 '13 at 17:58
  • @TheChaz2.0 Thanks a lot ! – A Googler Aug 11 '13 at 18:02
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After minar:

$2x^4+x^2+x+1= 2x^4+2x+x^2-x+1= 2x(x+1)(x^2-x+1)+x^2-x+1=(2x^2+2x+1)(x^2-x+1) $

Boris Novikov
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Just Simple Factorisation: $2a^4+a^2b^2+ab^3+b^4$ $=2a^4+2ab^3+a^2b^2−ab^3+b^4$ $=2a(a^3+b^3)+a^2b^2−ab^3+b^4$ $=2a(a+b)(a^2-ab+b^2)+b^2(a^2-ab+b^2)$ $=(a^2-ab+b^2)(2a^2+2ab+b^2)$

Rahul Chakwate
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