How do you use the rational roots theorem to find all possible zeros of #h(x) = 2x^4-5x^3+3x^2+4x-6#?

1 Answer
George C.
Jun 24, 2016

Zeros: #x=-1#, #x=3/2# and #x=1+-i#

Explanation:

#h(x) = 2x^4-5x^3+3x^2+4x-6#

By the rational roots theorem, any rational zeros of #h(x)# are expressible in the form #p, q# for integers #p, q# with #p# a divisor of the constant term #-6# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2#, #+-1#, #+-3/2#, #+-2#, #+-3#, #+-6#

Trying each in turn, we first find:

#f(-1) = 2+5+3-4-6 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#2x^4-5x^3+3x^2+4x-6#

#=(x+1)(2x^3-7x^2+10x-6)#

Then substituting #x=3/2# in the remaining cubic, we find:

#2x^3-7x^2+10x-6 = 2(27/8)-7(9/4)+10(3/2)-6#

#=27/4-63/4+60/4-24/4 = 0#

So #x=3/2# is a zero and #(2x-3)# is a factor:

#2x^3-7x^2+10x-6 = (2x-3)(x^2-2x+2)#

The remaining quadratic has negative discriminant, but we can find its Complex zeros by completing the square:

#x^2-2x+2 = x^2-2x+1+1#

#= (x-1)^2-i^2#

#= (x-1-i)(x-1+i)#

Hence zeros #x=1+-i#

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