How do you list all possible roots and find all factors and zeroes of 4x3−9x2+6x−1?

1 Answer
George C.
Jul 3, 2016

4x3−9x2+6x−1=(4x−1)(x−1)(x−1)

with zeros x=1 with multiplicity 2 and x=14.

Explanation:

f(x)=4x3−9x2+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 4x3−9x2+6x−1 must be expressible in the form pq 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:

±14,±12,±1

If we evaluate f(14) we find:

f(14)=464−916+64−1=1−9+24−1616=0

So x=14 is a zero and (4x−1) is a factor:

4x3−9x2+6x−1

=(4x−1)(x2−2x+1)

=(4x−1)(x−1)2

Hence we have zeros:

x=14

x=1 with multiplicity 2


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:

4x3−9x2+6x−1=(x−1)(4x2−5x+1)

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

4x2−5x+1=(x−1)(4x−1)

Putting it together:

4x3−9x2+6x−1=(x−1)(x−1)(4x−1)

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