How do you use the rational roots theorem to find all possible zeros of #f(x)=2x^3+x^2-13x+6#?

1 Answer
George C.
Mar 23, 2016

#2x^3+x^2-13x+6 = (2x-1)(x+3)(x-2)#

Explanation:

#f(x) = 2x^3+x^2-13x+6#

By the rational roots theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p# and #q# where #p# is 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#

Let us try each in turn:

#f(1/2) = 1/4+1/4-13/2+6 = 0#

So #x=1/2# is a zero and #(2x-1)# a factor:

#2x^3+x^2-13x+6 = (2x-1)(x^2+x-6)#

We could continue simply trying the other possible zeros, but it is quicker to note that #3xx2 = 6# and #3-2 = 1#, so:

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

Putting it all together, we find:

#2x^3+x^2-13x+6 = (2x-1)(x+3)(x-2)#

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