How do you factor #2x ^ { 2} + 3x y + 5y ^ { 2}#?

1 Answer
George C.
Oct 23, 2017

#2x^2+3xy+5y^2 = 1/8(4x+(3-sqrt(31)i)y)(4x+(3+sqrt(31)i)y)#

Explanation:

Given:

#2x^2+3xy+5y^2#

Note that this homogeneous quadratic in #x# and #y# is of the form:

#ax^2+bxy+cy^2#

with #a=2#, #b=3# and #c=5#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac#

#color(white)(Delta) = color(blue)(3)^2-4(color(blue)(2))(color(blue)(5))#

#color(white)(Delta) = 9-40#

#color(white)(Delta) = -31#

Since #Delta < 0# this quadratic has no linear factors with real coefficients.

It is still possible to factor it, but only with non-real complex coefficients.

We find:

#8(2x^2+3xy+5y^2) = 16x^2+24xy+40y^2#

#color(white)(8(2x^2+3xy+5y^2)) = (4x)^2+2(4x)(3y)+(3y)^2+31y^2#

#color(white)(8(2x^2+3xy+5y^2)) = (4x+3y)^2+(sqrt(31)y)^2#

#color(white)(8(2x^2+3xy+5y^2)) = (4x+3y)^2-(sqrt(31)iy)^2#

#color(white)(8(2x^2+3xy+5y^2)) = ((4x+3y)-sqrt(31)iy)((4x+3y)+sqrt(31)iy)#

#color(white)(8(2x^2+3xy+5y^2)) = (4x+(3-sqrt(31)i)y)(4x+(3+sqrt(31)i)y)#

So:

#2x^2+3xy+5y^2 = 1/8(4x+(3-sqrt(31)i)y)(4x+(3+sqrt(31)i)y)#

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