Question

How do you factor completely `P(x)=x^3-6x^2+11x-6`?

Answer 1

`x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)`

Explanation:

Given:

`P(x) = x^3-6x^2+11x-6`

Note that linear factors correspond to zeros. That is, `x=a` is a zero if and only if `(x-a)` is a factor.

`color(white)()`
Rational Roots Theorem

By the Rational Roots Theorem, any rational zeros of `P(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-6` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros of `P(x)` are:

`+-1, +-2, +-3, +-6`

`color(white)()`
Descartes' Rule of Signs

Note that the pattern of the signs of the coefficients of `P(x)` is `+ - + -`. With `3` changes, this means that `P(x)` has `3` or `1` positive real zeros.

The pattern of the signs of the coefficients of `P(-x)` is `- - - -`. With no changes, this means that `P(x)` has no negative real zeros.

So, in combination with the Rational Roots Theorem, we can deduce that the only possible rational zeros of `P(x)` are:

`1, 2, 3, 6`

We could try each of these values in turn and find that:

`P(1) = P(2) = P(3) = 0`

Hence `P(x) = (x-1)(x-2)(x-3)`

but here's another way...

`color(white)()`
Sum of coefficients shortcut

Note that the sum of the coefficients of `P(x)` is zero. That is:

`1-6+11-6 = 0`

Hence we can deduce that `x=1` is a zero of `P(x)` and `(x-1)` a factor:

`x^3-6x^2+11x-6 = (x-1)(x^2-5x+6)`

Then to factor the remaining quadratic note that `5 = 2+3` and `6 = 2*3`, and hence:

`x^2-5x+6 = (x-2)(x-3)`

Either way we arrive at our result:

`x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)`