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

Question

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!

Answer 1

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

Answer 2

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 $10$.

Answer 3

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 $q$.

Answer 4

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.....

Answer 5

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.