How do you factor #a^4+4b^4# ?

1 Answer
Jan 20, 2017

Answer:

#a^4+4b^4 = (a^2-2ab+2b^2)(a^2+2ab+2b^2)#

Explanation:

This polynomial factors nicely into two quadratic polynomials with integer coefficients:

#a^4+4b^4 = (a^2-2ab+2b^2)(a^2+2ab+2b^2)#

These quadratic factors have no simpler linear factors with Real coefficients. To see that, you can check their discriminants:

#Delta_(a^2-2ab+2b^2) = (-2)^2-4(1)(2) = 4-8 = -4#

#Delta_(a^2+2ab+2b^2) = 2^2-4(1)(2) = 4-8 = -4#

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Footnotes

#a^4+4b^4# is an example of a homogeneous polynomial in two variables, #a# and #b#. That is, all of the terms are of the same degree, namely #4# in this example.

Factoring homogeneous polynomials is very similar to factoring a corresponding polynomial in one variable.

In our example, we could let #t = a/b# and look at how to factor:

#t^4+4 = (t^2-2t+2)(t^2+2t+2)#

Then we could multiply this through by #b^4# to get our original factorisation.

Similarly, when we look at #a^2-2ab+2b^2#, the corresponding polynomial in #a/b# formed by dividing it by #b^2# can be expressed as:

#t^2-2t+2#

When you have a quadratic of the form #At^2+Bt+C# in one variable you are probably familiar with finding the discriminant #Delta = B^2-4AC#. The sign of the discriminant #Delta# allows us to determine whether the quadratic in #t# has Real zeros and therefore linear factors with Real coefficients.

Faced with #Aa^2+Bab+Cb^2# the same discriminant tells us whether this factors into linear factors with Real coefficients.