How do you use the rational root theorem to list all possible roots for #5x^3-11x^2+7x-1=0#?

1 Answer
George C.
Nov 21, 2016

The "possible" rational roots are: #+-1/5, +-1#

The actual zeros are: #1, 1, -1/5#

Explanation:

#f(x) = 5x^3-11x^2+7x-1#

By the rational root theorem, the only possible rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-1# and #q# a divisor of the coefficient #5# of the leading term.

That means that the only possible rational zeros are:

#+-1/5#, #+-1#

There could be other zeros, but they will be irrational or non-Real Complex. The rational root theorem only tells us possible rational zeros.

Note that the sum of the coefficients of #f(x)# is #0#. That is:

#5-11+7-1 = 0#

So #f(1) = 0#, #x = 1# is a zero and #(x-1)# a factor:

#5x^3-11x^2+7x-1 = (x-1)(5x^2-6x+1)#

The same is true of the remaining quadratic:

#5 - 6 + 1 = 0#

So we have another zero #x=1# and factor #(x-1)#:

#5x^2-6x+1 = (x-1)(5x+1)#

The remaining factor #(5x+1)# corresponds to zero #x=-1/5#.

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