<s>[INST] 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?
 [/INST] 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.

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