An algebra problem by Keerthi Reddy

Let the roots of $x^3-12x^2+39x-28$ be $a$ , $b$ and $c$ , using Vieta's formulas, we have:

$\begin{cases} a+b+c = 12 \\ ab + bc + ca = 39 \\ abc = 28\end{cases}$

From the options, we note that the roots should be $(1,4,7)$ and the factors are $\boxed{(x-1)(x-4)(x-7)}$ .

$\begin{aligned} x^3 - 12x^2 + 39x - 28 & = x^3 - (x^2 + 11x^2) + (11x + 28x) - 28\\ & = x^3 - x^2 - 11x^2 + 11x + 28x - 28\\ & = x^2(x - 1) - 11x(x - 1) + 28(x - 1)\\ & = (x - 1)(x^2 - 11x + 28)\\ & = (x - 1)(x^2 - 4x - 7x + 28)\\ & = (x - 1)\left[x(x - 4) - 7(x - 4)\right]\\ & = \boxed{(x - 1)(x - 4)(x - 7)} \end{aligned}$