Question

Determine the completely factored form of `f(x) = 12x^3 - 44x^2 + 49x - 15`?

Answer 1

`12x^3-44x^2+49x-15 = (2x-3)(3x-5)(2x-1)`

Explanation:

`f(x) = 12x^3-44x^2+49x-15`

By the rational roots theorem, any rational zero of `f(x)` is expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-15` and `q` a divisor of the coefficient `12` of the leading term.

In addition, note that the pattern of signs of coefficients of `f(x)` is `+ - + -` with `3` sign changes, while those of `f(-x)` are in the pattern `- - - -` with `0` sign changes. So by Descartes' Rule of Signs, `f(x)` has `3` or `1` positive Real zero and no negative Real zeros.

Hence the only possible rational zeros are:

`1/12, 1/6, 1/4, 1/3, 5/12, 1/2, 3/4, 5/6, 1, 5/4, 3/2, 5/3, 5/2, 3, 5, 15`

We could simply try each of these in turn, but if you are allowed, we can speed up the process of finding the zeros as follows:

Look at the derivative and find out where it is `0`, indicating a local maximum or minimum...

`f'(x) = 36x^2-88x+49`

`color(white)(f'(x)) = (6x-22/3)^2-484/9+49`

`color(white)(f'(x)) = (6x-22/3)^2-(sqrt(43)/3)^2`

`color(white)(f'(x)) = (6x-22/3-sqrt(43)/3)(6x-22/3+sqrt(43)/3)`

Hence zero at `x = 1/6(22/3+-sqrt(43)/3) = 1/18(22+-sqrt(43))`

`sqrt(43) ~~ 6.5`

So the local maximum and minimum are approximately at:

`1/18(22-6.5) = 31/36 ~~ 5/6` and `1/18(22+6.5) = 19/12 ~~ 3/2`

`f(5/6) = 20/9`

`f(3/2) = 0`

Hmmm - a zero near where we expect the minimum. Let's look at the other rational possibility nearby...

`f(5/3) = 0`

So `x=3/2` and `x=5/3` are zeros with corresponding factors `(2x-3)` and `(3x-5)`.

Looking at the coefficient of the leading term and constant term, the remaining factor (which must be rational) must be `(2x-1)`

So:

`12x^3-44x^2+49x-15 = (2x-3)(3x-5)(2x-1)`

graph{12x^3-44x^2+49x-15 [-0.448, 2.052, -1.26, 2.49]}