What are all the possible rational zeros for #f(x)=4x^3-9x^2+6x-1# and how do you find all zeros?

1 Answer
Feb 19, 2017

Answer:

#f(x)# has "possible" zeros:

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

and actual zeros:

#1/4, 1, 1#

Explanation:

Given:

#f(x) = 4x^3-9x^2+6x-1#

By the rational roots theorem, any 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 #4# of the leading term.

That means that the only possible rational zeros are:

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

Here's a non-standard way of finding the zeros:

I notice that the coefficients are all fairly small and the signs alternate.

Taking them in reverse order and stringing them together we get:

#1694 = 2*7*11*11 = 14*11*11#

Hence we find:

#t^3+6t^2+9t+4 = (t+4)(t+1)(t+1)#

Then putting #t = -1/x# we find:

#4x^3-9x^2+6x-1 = (4x-1)(x-1)(x-1)#

So #f(x)# has zeros #1/4, 1, 1#