Question

How do you factor `x^3-12x^2+41x-42`?

Answer 1

`x^3-12x^2+41x-42 = (x-2)(x-3)(x-7)`

Explanation:

Given:

`f(x) = x^3-12x^2+41x-42`

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 `-42` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1, +-2, +-3, +-6, +-7, +-14, +-21, +-42`

In addition, note that the pattern of signs of the coefficients of `f(x)` is `+ - + -`. With `3` changes of signs, Descartes' Rule of Signs tells us that `f(x)` has `3` or `1` positive real zeros. Also, the signs of the coefficients of `f(-x)` are in the pattern `- - - -`. With no changes of sign, Descartes' Rule of Signs tells us that `f(x)` has no negative real zeros.

So the only possible rational zeros are:

`1, 2, 3, 6, 7, 14, 21, 42`

We find:

`f(2) = (color(blue)(2))^3-12(color(blue)(2))^2+41(color(blue)(2))-42`

`color(white)(f(2)) = 8-48+82-42 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^3-12x^2+41x-42 = (x-2)(x^2-10x+21)`

To factor the remaining quadratic note that `3+7 = 10` and `3*7 = 21`. So:

`x^2-10x+21 = (x-3)(x-7)`

Putting it all together:

`x^3-12x^2+41x-42 = (x-2)(x-3)(x-7)`