Rearranging the polynomial $x^3-23x^2+142x-120$ prior to factoring it

Question

In the example 15: They are saying that,

$$x^3-23x^2+142x-120 = x^3-x^2-22x^2+22x+120x-120$$

From where did $22x^2$ and $22x$ come and also $120x$.

Please help me clear my confusion.

Answer 1

Note that $$-23x^2=-x^2-22x^2$$ and that $$142x=22x+120x.$$

So, we have

$$x^3\color{red}{-23x^2}+\color{blue}{142x}-120=x^3\color{red}{-x^2-22x^2}+\color{blue}{22x+120x}-120.$$

P.S. If you are asking the purpose (or background) of this rearrangement, then note that they noticed that $$142-120=22,\ \ 23-22=1$$ and that the coefficients can be $1,-1,-22,22,120,-120$ from the left. This is of course to factor as $$x^2(x-1)-22x(x-1)+120(x-1)=(x-1)(x^2-22x+120).$$

Answer 2

They are breaking up $-23x^2$ as $-x^2-22x^2$, and they're breaking up $142x$ as $120x+22x$. The purpose of this is to apply factoring by grouping in the next step.

Answer 3

First of all note that's an identity (the $=$ is true).

Now, they decide to rewrite the original polynomial in that fashion in order to actually use partial factorisation (they are going to "collect" $x-1$ every two addends). It's just a trick you can use, nothing really deep.