How do you list all possible roots and find all factors and zeroes of #4x^3-9x^2+6x-1#?

1 Answer
Jul 3, 2016

Answer:

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

with zeros #x=1# with multiplicity #2# and #x=1/4#.

Explanation:

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

I notice that the question asks for possible roots, so you are probably expected to make use of the rational root theorem first:

Since this cubic is given in standard form (with descending powers of #x#) and has integer coefficients, the rational root theorem can be applied:

Any rational zeros of #4x^3-9x^2+6x-1# must be 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#

If we evaluate #f(1/4)# we find:

#f(1/4) = 4/64-9/16+6/4-1 = (1-9+24-16)/16 = 0#

So #x=1/4# is a zero and #(4x-1)# is a factor:

#4x^3-9x^2+6x-1#

#= (4x-1)(x^2-2x+1)#

#= (4x-1)(x-1)^2#

Hence we have zeros:

#x=1/4#

#x=1# with multiplicity #2#

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Footnote

If the question did not mention "possible" roots, then I would have found the solution by looking at the sum of the coefficients first:

Note that #4-9+6-1 = 0#. Hence #f(1) = 0#, #x=1# is a zero and #(x-1)# a factor:

#4x^3-9x^2+6x-1 = (x-1)(4x^2-5x+1)#

Then note that #4-5+1 = 0#. Hence #(x-1)# is a factor again:

#4x^2-5x+1 = (x-1)(4x-1)#

Putting it together:

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