# Long division of polynomials: $(2x^4 - 5x^3 - 15x^2 + 10x +8) \div (x^2-x-2)$

I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics". I'm currently stuck on an aspect of long division of polynomials, when the denominator itself is a polynomial. So, I know how to do long division when the deniminator is something like $$(x+1)$$, but when it's a polynomial, I'm not sure what the mechanics are, like in the below example:

$$(2x^4 - 5x^3 - 15x^2 + 10x +8) \div (x^2-x-2)$$

Can anybody shed some light here, please? Thank you!

• But $x+1$ is also a polynomial... Apr 4, 2019 at 12:27
• The division works the same way for quadratics as the divisor, however you could also factor $x^2-x-2=(x-2)(x+1)$ and then divide your quartic by $x-2$ first then divide this result by $x+1$.
– Dave
Apr 4, 2019 at 12:30

It can be handled in a similar way to normal long division: $$\require{enclose} \begin{array}{r} \color{#C00}{2x^2}\ \color{#090}{-3x}\color{#00F}{-14}\phantom{{}+10x+8}\quad\\[-4pt] x^2-x-2\enclose{longdiv}{2x^4 - 5x^3 - 15x^2 + 10x\ +8}\\[-4pt] \underline{\color{#C00}{2x^4-2x^3-\,4x^2}}\phantom{10x+8}\quad\ \ \ \\[-2pt] -3x^3-11x^2\phantom{10x+8}\quad\ \,\,\\[-3pt] \underline{\color{#090}{-3x^3+\ \ 3x^2+\ \ 6x}}\phantom{{}+8}\ \ \\[-2pt] -14x^2+\ 4x\phantom{{}+8}\ \ \,\\[-3pt] \underline{\color{#00F}{-14x^2+14x+28}}\\[-3pt] -10x-20 \end{array}$$

• +1, good use of color. Apr 6, 2019 at 2:59

Hint. The leading term in the quotient will have to be $$2x^2$$ in order for the product with the denominator to match the leading term in the numerator. Then subtract and continue inductively - just as with the classic (in the U.S.) algorithm for long division of integers in base $$10$$.

Long division of a polynomial $$f$$ by another polynomial $$g$$ results in polynomials $$q$$ and $$r$$ with $$\deg r<\deg g$$ such that $$f=qg+r$$. In other words, it results in a polynomial $$q$$ such that $$\deg(f-qg)<\deg g.$$ If $$\deg f<\deg g$$ then you are immediately done with $$q=0$$ and $$r=f$$. Otherwise you can subtract a multiple of $$g$$ from $$f$$ to get a polynomial of lower degree. In your particular example $$f-qg=(2x^4-5x^3-15x^2+10x+8)-q(x^2-x-2),$$ so for this to be a polynomial of lower degree than $$f$$, the leading terms must cancel. This means that the leading term of $$q$$ must be $$2x^2$$. We see that $$f-2x^2g=(2x^4-5x^3-15x^2+10x+8)-2x^2(x^2-x-2)=-3x^3+11x^2+10x+8.$$ Now repeating this process with the resulting polynomial $$f_1:=f-2x^2g=-3x^3+11x^2+10x+8,$$ yields the next term of $$q$$, and repeating again yields the last term of $$q$$.

univariate (1 variable, instead of more than 1) polynomials are a generalization of base x numbers. You can almost treat this as a base 16 division with borrowing prior (to allow subtraction in) and remainder. $$\begin{eqnarray}1C1\quad \text{ rem }3A\\\hspace{-250px}EE\big)\overline{1A1A8}\end{eqnarray}$$

rewritten as polynomials that's $$x^2+12x+1$$ remainder $$3x+10$$ Undoing most if not all of the borrowing we did before, we get $$2x^2-3x-15$$ which we can check gives $$2x^4-5x^3-14x^2+21x+30$$ We are over by $$x^2+11x+22$$ adding the remainder gives $$x^2+14x+32$$ which then leaves 15x+34 on subtraction of the divisor, adding 1 to the multiplier giving :$$2x^2-3x-14$$ and remainder of $$15x+34$$. My working out was hard to align using hspace,and partially calculator assisted so I took it out.

Usually in these problems, I will first analyze the denominator polynomial $$g(x)=x^2-x-2$$. The roots of this polynomial are: $$g(x)=(x-2)(x+1)=0$$ giving, $$x=-1,2$$. Now the second thing I do is I check whether the numerator polynomial is having a common root. So we check $$f(x)=2x^4-5x^3-15x^2+10x+8$$, for $$x=-1,2$$.

$$f(-1)=-10$$ and $$f(2)=-40$$, so none of the roots are common telling that this division will lead to a remainder.

The next thing I would do is just take $$(x-2)$$ common from $$f(x)$$ like this (take common whatever is remaining you just add to make it equal to $$f(x)$$):

$$f(x)=2x^3(x-2)-x^2(x-2)-17x(x-2)-24(x-2)-40$$ $$\frac{f(x)}{g(x)}=\frac{2x^3(x-2)-x^2(x-2)-17x(x-2)-24(x-2)-40}{(x-2)(x+1)}$$ $$\frac{f(x)}{g(x)}=\frac{2x^3-x^2-17x-24}{x+1}-\frac{40}{(x-2)(x+1)}$$ For the first part do the same thing again. Take $$(x+1)$$ common to get: $$2x^2(x+1)-3x(x+1)-14(x+1)-10$$ $$\frac{f(x)}{g(x)}=\frac{(2x^2-3x-14)(x+1)}{x+1}-\frac{10}{x+1}+\frac{40}{(x-2)(x+1)}$$

$$\frac{f(x)}{g(x)}=2x^2-3x-14-\frac{10}{x+1}-\frac{40}{(x-2)(x+1)}$$ This is the final answer for your division. You can also club the last two terms so that you get a common denominator function. $$\frac{f(x)}{g(x)}=2x^2-3x-14-\frac{10(x+2)}{(x-2)(x+1)}$$ You can also try synthetic division which is far more easier but for that you need to have a good hold on the concept of long division. Hope this helps.....

• The $-40$ in the top two equations becomes a $+40$ in the third (and beyond). This accounts for the difference in our remainders.
– robjohn
Apr 5, 2019 at 1:47