How to divide polynomials?

Question

The title says it all : How to divide polynomials?

I don't understand the method below taught in my school.
Would any of you mind explaining it or even better, suggest an alternative way to solve this?

Answer 1

This is an easier way to do polynomial division, at least I think it is easier.

Basically when we multiply polynomials (or any numbers really) we can sometimes do it in a table form, say if we wanted to multiply $x + 2$ and $x^2 + 3x + 4$ we might write a table like this: $$ \\ \begin{array} {c|c|c|c} multiply&x^2 & 3x &4 \\ \hline x &x^3 &3x^2 &4x \\ \hline 2 &2x^2 & 6x & 8 \end{array} $$

And from this we get that the product of $x + 2$ and $x^2 + 3x + 4$ is $x^3 + 5x^2 + 10x + 8$. Now we want to do this in reverse.

We know that the highest power term in the product ($2x^2$ in this case) must be the product of the highest powers in the factors. This might seem a bit strangely worded, but basically what I mean is that we know our quadratic (not trinomial) is the product of two linear (not binomial) factors, so our $2x^2$ term only comes from the product of the $x$ terms. So, we know when we do the table multiplication as above we have:

$$ \\ \begin{array} {c|c|c} divide& * & \\ \hline x &\color{Red}{2x^2} & \\ \hline 1 &** & \end{array} $$

And now we know what the entry in $*$ must be: $2x$, and then we get $**$ also so we have

$$ \\ \begin{array} {c|c|c} divide& \color{Red}{2x} & ** \\ \hline x &{2x^2} & * \\ \hline 1 & \color{Red}{2x}& \end{array} $$

And now, because in our product $2x^2 + 5x + 3$ we have $5x$, and the power of the term in $*$ is x, we must have that $2x + * = 5x$ so we get $ * = 3x$ so then $** = 3$ so our table is :

$$ \\ \begin{array} {c|c|c} divide& {2x} & \color{red}{3} \\ \hline x &{2x^2} & \color{red}{3x} \\ \hline 1 & {2x}& * \end{array} $$

And lastle we see * = 3 which is what we want, so we get that $(x+1)(2x+3) = 2x^2 + 5x + 3$ so $\frac{2x^2 + 5x + 3} {x + 1} = 2x + 3$.

This works in general for any order polynomial, and if one polynomial doesn't divide the other, you can easily get the remainder from this algorithm too.

Answer 2

Here is another method that you asked for. Its main advantage is that you can easily do this in a text editor, which helps immensely in accurately tracking your work. It is completely done with algebraic manipulation.

So, with $A$ as the highest term in numerator and $B$ as the sum of all remaining terms. And $C$ as the highest term in denominator, and $D$ is the remaining terms. The goal is to eliminate $A$ and get an answer in terms of $m + \frac{r}{C + D}$. In other words, $m$ is a multiple of the denominator, and $r$ is the remainder. So you start with two things being divided.

Here we justify an algebraic procedure. Imagine some example: $$ \color{red}{A = 4x^2} $$

$$ \color{green}{B = 3x + 5} $$

$$ \color{blue}{C = 2x^2} $$

$$ \color{purple}{D = 9x - 7} $$

So the polynomial being divided is like this, at a high level.
So ignore their values and see what happens when we perform one step: $$ \frac{A + B}{C + D} = $$

And note that you can add and subtract a ratio of the highest terms $\frac{A}{C}$ without changing the answer.

$$ \frac{A}{C} - \frac{A}{C} + \frac{A + B}{C + D} = $$

And you can multiply and divide by the denominator $C + D$ without changing the answer as well. Specifically, you can do this to the subtracted term. It just happens that $A$, which represents the highest term will always cancel out when you simplify this. This is actually the only thing you need to remember. You can derive everything else with this.

$$ \frac{A}{C} - \frac{A}{C}*\frac{C+D}{C+D} + \frac{A + B}{C + D} = $$

If you start there, you can usually jump right to that step in your head. So now, we can figure out what it simplifies to, just to see what happens with $A$:

$$ \frac{A}{C} + \frac{-AC -AD}{CC + CD} + \frac{AC + BC}{CC + DC} = $$

$$ \frac{A}{C} + \frac{BC - AD}{C(C + D)} $$

So, let's apply it to the example values:

$$ \frac{\color{red}{4x^2} + \color{green}{(3x + 5)}}{\color{blue}{2x^2}+ \color{purple}{(9x - 7)}} = $$

$$ \frac{\color{red}{4x^2}}{\color{blue}{2x^2}} + \frac{(\color{green}{3x + 5})(\color{blue}{2x^2}) - (\color{red}{4x^2})(\color{purple}{9x-7})}{(\color{blue}{2x^2})(\color{blue}{2x^2}+ \color{purple}{(9x - 7)})} = $$

$$ \frac{4x^2}{2x^2} + \frac{6x^3 + 10x^2 - 36x^3 + 28x^2}{(2x^2)(2x^2+ (9x - 7))} = $$

$$ \frac{4x^2}{2x^2} + \frac{3x + 5 - 18x + 14}{2x^2+ (9x - 7)} = $$

$$ 2 + \frac{-15x + 19}{2x^2+ (9x - 7)} $$

So, the current answer is $2$ plus remainder $(-15x+19)$. If $\frac{-15x}{2x^2} $ has a positive $x$ exponent, then we apply this method to the remainder to get more terms. Let's check the answer:

$$ 2(2x^2+ (9x - 7)) + (-15x + 19) = $$

$$ (4x^2+ 18x - 14) + (-15x + 19) = $$

$$ \color{red}{4x^2} + \color{green}{3x +5} $$

Which is our numerator $\color{red}{A} + \color{green}{B}$.

Let's apply this to a different polynomial that requires more than one reduction:

$$ \frac{\color{red}{5x^2} + \color{green}{2x + 3}}{\color{blue}{x} + \color{purple}{1}} = $$

$$ \frac{5x^2}{x} + \frac{(2x + 3)x - 5x^2}{x(x + 1)} = $$

$$ 5x + \frac{2x + 3 - 5x}{x + 1} = $$

$$ 5x + \frac{\color{red}{-3x} + \color{green}{3}}{\color{blue}{x} + \color{purple}{1}} = $$

$$ 5x - 3 + \frac{3x + 3x}{x(x + 1)} = $$

$$ 5x - 3 + \frac{6}{x + 1} = $$

So, let's check that the answer is $5x - 3$ with remainder $6$.

$$ (5x - 3)(x + 1) + 6 = 5x^2 + 5x - 3x - 3 + 6 = 5x^2 + 2x + 3 $$

Answer 3

It's regular long division with the divisor on the right instead of on the left. And in long division you always...subtract. But their notation is wrong. The brackets should not be there, OR that negative inside the brackets should be plus. So there is a notational issue here. Try to do the division yourself to see, I did it just now...