Hi,
I just refreshed on long division with polynomials, and I took a look at synthetic division. It's pretty cool, but I wonder why it works. I could kinda see how just coefficients can do the trick, but I am not very sure...
Thanks!
-Masoug
Hi,
I just refreshed on long division with polynomials, and I took a look at synthetic division. It's pretty cool, but I wonder why it works. I could kinda see how just coefficients can do the trick, but I am not very sure...
Thanks!
-Masoug
Its really not that different form regular long division. Of course, it only applies to division by "x- a" while with regular division, you can divide any polynomial by any polynomial.
To divide, say, $\displaystyle 2x^3- 3x^2+ 2x+ 1$ by x- 1, you would think "$\displaystyle x$ divides $\displaystyle 2x^3$ $\displaystyle 2x^2$ times". So your first term in the quotient is "2x" and you multiply x- 1 by that: $\displaystyle 2x^3- 2x^2$ and subtract: that leaves $\displaystyle x^2+ 2x+ 1$. Notice that the coefficient in the quotient, 2, is exactly the coefficent of $\displaystyle 2x^3$ in the dividend precisely because the coefficient of x in the divisor is 1.
Now divide $\displaystyle x^2+ 2x+ 1$ by x- 1. Again, because the coefficient of x in the divisor is 1, we get precisely $\displaystyle "1"x$ as quotient. And then we subtract $\displaystyle x(x- 1)= x^2- x$ from $\displaystyle x^2+ 2x+ 1$ to get $\displaystyle 3x+ 1$. Finally, of course, x divides into 3x 3 times so the entire quotient is $\displaystyle 2x^2+ x+ 3$ and [maht]3x+ 1- 3(x-1)= 3x+ 1- 3x+ 3= 4[/tex] is the remainder.
Now look at what we do with "synthetic division". Since we are [b]always[b] dividing by x- a the only thing that changes from one division to another is the "a" that's all we need to write. Because we always have the dividend written in descending powers of x, we don't need to write the "x"s, just the coefficients. And, because the coefficient of x in "x- a" is always 1, the coefficient in the quotient is always the coefficient of the first term in the dividend- we can just bring that down. Oh, and because it is alway -a, we can reverse the sign on a and add rather than subtract!
$\displaystyle \begin{bmatrix}1 | & 2 & -3 & 2 & 1 \\ \underline{|}& \underline{_} & \underline{2} & \underline{-1} & \underline{3} \\ & 2 & -1 & 3 & |-2\end{bmatrix}$
Compare that with the "long division" and you will see what I mean.