I was reading the free sample of the book: A Synopsis of Elementary Results in Pure and Applied Mathematics, when I came up this:
And no matter how much I think about it, and look at it, I can't see how the author followed through. The instructions also kind of confuse me.
I understand how he got \begin{align*} & 1-2-6+4+13+6\\ 3\end{align*} \begin{align*}-----------\\ 3-6-18+12+39+18\end{align*} But I'm not sure what the $1$ near the outer-edge means, nor where the author got $-3-4+6+12+5$.
The organization of the work in the example, particularly the placement of the results, is pretty misleading, although if you had to do hundreds of such problems by hand, it is a pretty efficient organization to do once you understand it.
Let me fill in the blanks by filling in the powers of $x$ and placing things in a logical order.
Start with $3x^5 -10 x^3 + 15x + 8$, the top line on the right. We want to get rid of the $x^5$ term, and what we have to work with is $x^5-2x^4-6x^3+4x^2+13x+6$ (which is placed as the top row on the left). Well, the $x^5$ terms don't match, but they would if we multiplied the row on the left by $3$ (which we have written on the right, and also as the second row on the left to do the multiplication.
This multiplication gives $$ (3) (x^5-2x^4-6x^3+4x^2+13x+6) = 3x^5-6x^4-18x^3+12x^2+39x+18$$ which is the third row on the left. And we carry that over to the second row on the right, and subtract to get rid of the $x^5$ terms. That leaves $6x^4 + 8x^3 -12x^2 -24 x-10$ which is the third row on the left. Since we prefer to work with an expression with a leading coefficient of $1$, we can factor out a $2$, yielding the fourth row $2(3x^4 + 4x^3 -6x^2 -12 x-5)$.
For your understanding, it is important to see what we have done so far: $$3x^5 -10 x^3 + 15x + 8 = 3(x^5-2x^4-6x^3+4x^2+13x+6)+2(3x^4 + 4x^3 -6x^2 -12 x-5)$$ Now we need to get rid of the $3x^5$ appearing in the third row of the left column (or in the right side of the last equation presented). Luckily, we have $(3x^4 + 4x^3 -6x^2 -12 x-5)$ available, so we multiply that by $1$ (written on the outside of row $3$ left) time $x$, and subtract (using row $4$ on the left) obtaining the fifth row on the left: $$ 3(x^5-2x^4-6x^3+4x^2+13x+6) - x(3x^4 + 4x^3 -6x^2 -12 x-5) = -10x^4 -12x^3 + 22x^2 + 44x + 18$$ and we notice all those coefficients are even so we can factor out a $2$ obtaining $(-5x^4 -6x^3 + 11x^2 + 22x + 9)$, written as the sixth row on the left. So far we have $$3x^5 -10 x^3 + 15x + 8 = 3(x(3x^4 + 4x^3 -6x^2 -12 x-5)+(-10x^4 -12x^3 + 22x^2 + 44x + 18))+2(3x^4 + 4x^3 -6x^2 -12 x-5)\\= (3x+2)(3x^4 + 4x^3 -6x^2 -12 x-5)+3(2)(-5x^4 -6x^3 + 11x^2 + 22x + 9))$$
Now here is where the organization shown is completely abysmal and hard to follow:
We need to get rid of $x^4$ and we have available the expressions $(-5x^4 -6x^3 + 11x^2 + 22x + 9)$ and $(3x^4 + 4x^3 -6x^2 -12 x-5)$. While it would make sense to multiply one of these through by a fraction (e.g., $\frac53$) the author chooses to work with integers, which is perhaps easier to do by hand. So he multiplies the former by $3$, obtaining $(-15x^4 -18x^3 + 33x^2 + 66x + 27)$ (which you will recognize as line $7$ on the left), and he multiplies the latter by $5$, obtaining $(15x^4 + 20x^3 -30x^2 -60 x-25)$ (row $8$). He adds these to get $$ 2x^3 + 6x^2 +6x+2 = 2(x^3 + 3x^2 +3x+1) $$ on the ninth and tenth rows on the left.
He then multiplies that by $-3$ (row $5$ on the right) and adds, to get rid of the $x^4$ term in $(3x^4 + 4x^3 -6x^2 -12 x-5)$, obtaining $-5x^3 - 15x^2 - 15x - 5$ which he notices is $(-5)((x^3 + 3x^2 +3x+1)$.
If you do the exercise of substituting each of these nested expressions, you will find that what he has shown is that $(x^3 + 3x^2 +3x+1)$ divides each of the original expressions, and that no polynomial of higher degree does so.