I want to get faster at multiplying and dividing polynomials. Is there a way I can do it faster? For multiplying I just do it normally. But for division I have to use the box method.
$\endgroup$ 52 Answers
$\begingroup$Here is a way I like to do division of polynomials that has the advantage that it doesn't require writing anything down except the final answer; I find that this is convenient and faster than setting everything up. Here's an example:
Suppose I want to divide $2x^4+3x^3+x^2+1$ by $x^2+x+1$. Maybe I start by writing (though you can omit all the writing except the answer if you want) something like $2x^4+3x^3+x^2+1 = (x^2+x+1)($, where I'm going to start writing my answer after that parenthesis. I think, what must the highest order term be? Well, whatever I multiply $x^2+x+1$ by, the highest degree term will be a multiple of that $x^2$. We need to end up with $2x^4$, so, I see I'll want to multiply the $x^2$ by $2x^2$, and so I write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2$. Now, I know whatever lower terms I add on to my answer, I've gotten the fourth power of the end result correct.
So, now I move on to the third power. We want our product $2x^4+3x^3+x^2+1$ to have a $3x^3$ term in it, and we know we want to multiply our $x^2+x+1$ by some multiple of $x$ to make this happen. You need to be careful though: looking at our $(x^2+x+1)(2x^2$, we see that we have already contributed to the $x^3$ term in the answer. In particular, when we multiply what we have so far, the squared $2x^2$ and the linear $x$ in $x^2+x+1$ multiply to the cubic term $2x^3$. So, we already have $2x^3$ of the $3x^3$ that we want, so we only need to get a $1x^3$ more: so we write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x$, as that new $x$ term combines with the $x^2$ to make the $x^3$.
Now we know that we've secured the correct fourth and third power of our final product, so it's time to work on the $x^2$ part. $2x^4+3x^3+x^2+1$ just has $x^2$ for that term, but we need to see how our previous terms we've written so far have affected that term. So looking at $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x$, we see two multiplications ended up making a contribution to that: the $1 * 2x^2$ and $x*x$, which make $3x^2$. So we want an $x^2$ in the end, but already have contributed $3x^2$, so we actually need to remove $2x^2$. So we write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2$, as that $-2$ will multiply by that $x^2$ to contribute what we want. (note in this step, we had to check two contributions that our previous terms did to the $x^2$ term. For longer polynomials, this step gets harder and harder, involving summing three, four, etc terms; the nice thing is that when doing all this adding, you don't need to keep much in your head at one time, as you can figure out what one previous thing you wrote contributed, and move on, keeping a running total. So even for large polynomials, if your arithmetic is fast and accurate, you can use this)
Now we've found the constant term, so next we need to find the part that didn't divide in nicely. Perhaps we write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2 + \frac{}{x^2+x+1}$, and we know on the top of that fraction we have a linear term, to deal with the linear and constant term of the product. We see $2x^4+3x^3+x^2+1$ has no linear term, but our $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2$ contributed to the linear term $x*1$ and $-2*x$, for a net of $-x$; so we need to add $x$ back in to cancel that. So we write $(x^2+x+1)(2x^2+x-2 + \frac{x}{x^2+x+1}$. Finally, the constant term of $2x^4+3x^3+x^2+1$ is $1$, but in $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2$ we already contributed $-2$ to the constant term, so we need to use the fraction to add a $3$, so we write $(x^2+x+1)(2x^2+x-2 + \frac{-x+3}{x^2+x+1})$.
For multiplication, you can similarly work out the product term by term by keeping running totals: it's easy to determine the constant term, two products will contribute to the linear term of the answer, three to the quadratic term, and so on. It's an easier process than the case of division, so I'll leave the details for you to ponder.
$\endgroup$ 3 $\begingroup$You’re asking about getting faster. The only trick I know is to practice a lot. What’s always used to slow me down was making mistakes and having to redo my work.
$\endgroup$ 1
