Simplify $bx(x-b) + ax(a-x) + ab(b-a)$ to $-(a-b)(a-x)(b-x)$

Question

I was doing some matrices problems when I got to this factoring $$bx(x-b) + ax(a-x) + ab(b-a)$$ I found this was the answer, but it was
$$-(a-b)(a-x)(b-x)$$ Doing some researches, I found this remarkable product $$(a–b)(a–c)(b–c) = ab(a–c) + bc(b–c) + ca(c–a)$$ but with no explanation at all of how to do this simplification.

Answer 1

$f(x)=bx(x-b)+ax(a-x)+ab(b-a)$ is a quadratic equation.

We can check that $a$ and $b$ are roots.

Hence we can write $f(x)=A(x-a)(x-b)$ for some $A$. By verifying the leading coefficient of $x^2$, we can conclude that $A=b-a$.

Answer 2

Since the target expression has factor $(a-b)$ showing up at the original expression as $(b-a)$, we have a clue how to get this started $\to$ \begin{align}bx(x-b) + ax(a-x) + ab\color{green}{(b - a)}\end{align} \begin{align}=bx^2-b^2x + a^2x-ax^2 + ab\color{green}{(b - a)} \end{align} \begin{align}=x^2\color{green}{(b - a)}-\color{green}{(b - a)}(b+a)x+ ab\color{green}{(b - a)}\end{align} \begin{align}=\color{green}{(b - a)}(x^2-(b+a)x+ ab)\end{align} \begin{align}=\color{green}{(b - a)}(x-a)(x-b) = -(a-b)(a-x)(b-x)\end{align}