Question

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

Answer 1

`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)`