How do you factor #5a^2-6a+5#?

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Answer 1 · 2015-09-27T20:18:03

#5a^2-6a+5 = 1/5(5a-3-4i)(5a-3+4i)#

To try to avoid confusion, I will factor #5x^2-6x+5#, which is of the form #ax^2+bx+c#, with #a = 5#, #b = -6# and #c = 5#.

This has discriminant #Delta# given by the formula:

#Delta = b^2 - 4ac = (-6)^2 - (4xx5xx5) = 36 - 100#

#= -64 = -8^2#

Since this is negative, #5x^2-6x+5 = 0# only has Complex roots, given by the quadratic formula:

#x = (-b +-sqrt(Delta))/(2a) = (6+-sqrt(-8^2))/(2xx5) = (6+-8i)/10 = (3+-4i)/5#

Hence:

#5x^2-6x+5 = 1/5(5x-3-4i)(5x-3+4i)#

So:

#5a^2-6a+5 = 1/5(5a-3-4i)(5a-3+4i)#