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?
 A: 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.
A: The algorithm is straightforward and is loosely similar to the usual long division of decimal numbers as follows:


*

*Find the quotient of the leading term in the numerator divided by the leading term in the denominator.  In your case, $3x^3/x^2 = 3x$.

*Multiply the quotient in step 1 by the denominator.  You should get $3x^3 - 3x^2 - 18x$.

*Subtract the numerator from the result in step 2.  You should have $3x^3 - x^2 - 13x - 13 - (3x^3 - 3x^2 - 18x) = 2x^2 + 5x - 13$.  This is the remainder from the first cycle.

*If the degree of the remainder in step 3 is less than the degree of the denominator, stop.  Otherwise, repeat steps 1-3 with the result from step 3 as your new numerator.  $2x^2/x^2 = 2$, so the next quotient is $2$.


Upon completion, you should have quotients $3x$, $2$, and a remainder of $7x-1$.
A: First, you see which monomial you must multiply $x^2-x-6$ by in order to get something close to $3x^2-x^2-13x-13$. Clearly, that would be $3x$: $3x(x^2-x-6)=3x^3-3x^2-18x$. Now, you subtract this from $3x^2-x^2-13x-13$:$$3x^2-x^2-13x-13-(3x^3-3x^2-18x)=2x^2+5x-13.$$Now, you start all over again. What is the monomial you must multiply $x^2-x-6$ by in order to get something close to $2x^2+5x-13$? Clearly, it is $2$: $2(x^2-x-6)=2x^2-2x-12$. Now, you subtract this from $2x^2+5x-13$:$$2x^2+5x-13-(2x^2-2x-12)=7x-1.$$Since you now have a polynomial whose degree is smaller than the degree of the polynomial that you are dividing by, you're done:


*

*the quotient is the sum of the monomials that you got: $3x+2$;

*that final polynomial ($7x-1$) is the remainder.

A: Polynomial long division is exactly same as long division of decimals.
With respect to your posed problem here are steps.
Step1: We have to divide $3x^3-x^2-\dots$ with $x^2$, so we first multiply the divisor $(x^2-x-6)$ by $3x$ to give $3x^3$. THen we subtract and copy the remaining terms.
Step2: Again now we have $2x^2+5x-13$ and we have to divide by $x^2-x-6$. So we multiply our divisor by $2$.
Finally, we get $3x+2$ as dividend and $7x-1$ as remainder. Now recall if we had  $\frac{17}{3}$ we would write it as $5+\frac{2}{3}$. Here $\bf{2}$ is remainder.
Similarly we write our result of long division as $$3x+2+\frac{7x-1}{x^2-x-6}$$

