How do you factor and find the zeroes for #g(x)= x^3 -2x^2 + 5x#?

1 Answer
George C.
May 30, 2015

#g(x) = x^3-2x^2+5x = x(x^2-2x+5)#

So when #x=0#, #g(x) = 0#

The quadratic #x^2-2x+5# is of the form #ax^2+bx+c# with #a=1#, #b=-2# and #c=5#. This has discriminant given by the formula:

#Delta = b^2-4ac = (-2)^2-(4xx1xx5) = 4 - 20 = -16#

Since #Delta < 0# the quadratic has no real roots. It has two distinct complex roots.

The only real zero of #g(x)# is #x=0#

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