How do you factor #3a ^ { 2} - a b + 2b ^ { 2}#?

1 Answer
George C.
May 4, 2017

Only possible to factor with Complex coefficients:

#3a^2-ab+2b^2 = 1/12(6a-(1+sqrt(23)i)b)(6a-(1-sqrt(23)i)b)#

Explanation:

Given:

#3a^2-ab+2b^2#

Note that all of the terms are of degree #2#, so factoring such a polynomial is similar to factoring the following polynomial in one variable with the same coefficients:

#3x^2-x+2#

Being in the form #ax^2+bx+c#, we find that this polynomial has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-1)^2-4(3)(2) = 1-24 = -23#

Since #Delta < 0# this quadratic has no real-valued zeros and no linear factors with real coefficients.

Similarly #3a^2-ab+2b^2# has no linear factors with real coefficients.

#color(white)()#
Complex solution

We can still factor the given polynomial by completing the square (or by using the quadratic formula) and allowing complex coefficients.

For example:

#12(3a^2-ab+2b^2) = 36a^2-12ab+24b^2#

#color(white)(12(3a^2-ab+2b^2)) = (6a)^2-2(6a)b+b^2+23b^2#

#color(white)(12(3a^2-ab+2b^2)) = (6a-b)^2-(sqrt(23)ib)^2#

#color(white)(12(3a^2-ab+2b^2)) = ((6a-b)-sqrt(23)ib)((6a-b)+sqrt(23)ib)#

#color(white)(12(3a^2-ab+2b^2)) = (6a-(1+sqrt(23)i)b)(6a-(1-sqrt(23)i)b)#

So:

#3a^2-ab+2b^2 = 1/12(6a-(1+sqrt(23)i)b)(6a-(1-sqrt(23)i)b)#

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