Long division of $\frac{3x^3-x^2-13x-13}{x^2-x-6}$ I'm self-studying from Stroud & Booth's amazing textbook "Engineering Mathematics", and am on the "Partial Fractions" chapter. As part of an exercise I need to do long division of two polynomial equations.
The problem is, long division of polynomials was never explained in the textbook. The long division I need to do is this one:
$$\frac{3x^3-x^2-13x-13}{x^2-x-6}$$
The book states that the result is:
$$3x+2+\frac{7x-1}{x^2-x-6}$$
But the mechanics of the long division are never explained. Can I please get some help with this one?

For long division, you essentially keep reducing the degree of the numerator until it's lower than the denominator.
In the example you gave, $\displaystyle \frac{3x^3 - x^2 - 13x - 13}{x^2 - x- 6}$, we first look at the leading term, $3x^3$. We now want to remove this from the numerator, so we multiply $x^2-x-6$ by $3x$ to obtain $3x^3 - 3x^2 - 18x$. We now write the numerator with this term in it: $$\displaystyle \frac{3x^3 - x^2 - 13x - 13}{x^2 - x- 6} = \frac{(3x^3 - 3x^2 - 18x) + 2x^2 + 5x - 13}{x^2 - x- 6} = 3x + \frac{2x^2 + 5x - 13}{x^2 - x- 6}.$$
We now repeat the process again; we remove the leading term $x^2$on the numerator by multiplying $x^2-x-6$ by $2$ to obtain $2x^2 - 2x - 12$, and then write the numerator with this term:
$$3x + \frac{2x^2 + 5x - 13}{x^2 - x- 6} = 3x + \frac{(2x^2 -2x - 12) + 7x - 1}{x^2 - x- 6} = 3x + 2 + \frac{7x-1}{x^2-x-6}.$$
The numerator now has lower degree than the denominator, and so we stop.
