What are all the zeroes of #g(x)= 2x^3 - 5x^2 + 4 #?

1 Answer
Aug 8, 2016

Answer:

#g(x)# has zeros #2# and #1/4+-sqrt(17)/4#

Explanation:

#g(x) = 2x^3-5x^2+4#

By the rational root theorem, any rational zeros of #g(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #4# and #q# a divisor of the coefficient #2# of the leading term.

That means the only possible rational zeros are:

#+-1/2, +-1, +-2, +-4#

We find:

#g(2) = 2(8)-5(4)+4 = 16-20+4 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#2x^3-5x^2+4 = (x-2)(2x^2-x-2)#

The remaining quadratic is in the form #ax^2+bx+c# with #a=2#, #b=-1# and #c=-2#. This has disrciminant #Delta# given by the formula:

#Delta = b^2-4ac = (-1)^2-4(2)(-2) = 1+16 = 17#

Since this is positive but not a perfect square, the remaining zeros are Real but irrational. They are given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#= (1+-sqrt(Delta))/4#

#= (1+-sqrt(17))/4#

#= 1/4+-sqrt(17)/4#