Algebra 2 students all over the country are being taught to use “synthetic division” to divide a polynomial by a linear expression. It goes something like this:
We need to do some algebraic manipulation next:
Here we have a factorization of the original polynomial, so the expression on the left must be the quotient we were seeking. But how exactly does this work???
A curious person might wonder: where does 3/2 come from? And why, through repeated cycles of multiplying and adding, does this algorithm produce the quotient polynomial we were seeking? It seems to operate by magic!
Let’s try to take the mystery out of synthetic division. To do so, let’s go back to an old friend: multiplication. We’ll begin by multiplying (2x^2 + 5x -1) by (2x – 3). We choose a box with an appropriate number of “rows” and “columns.” With a small stretch of the imagination, we allow for the number of rows to be a variable (e.g. 2x rows), and we also allow rows or columns to have negative values (e.g. -1 columns). So we’re looking at this picture:
We proceed by finding the area of each box:
Next, we notice that terms with like exponents lie on the diagonals of the box. (To see why this is true, try drawing this diagram with two standard polynomials where all the coefficients are just 1.)
Adding like terms gives us an expression for the total area of the box, which is the product we were seeking.
Grand! We have succeeded at multiplying two polynomials. Now it’s time to put on our work gloves and attempt a division problem. Why don’t we try doing the same problem backwards? That would certainly give us added confidence in our solution. Here goes: let’s divide 2x^3 + 7x^2 – 17x + 3 by 2x – 3.
Now — how to get started? Well, since the like terms are on the diagonals, it must be true that the corner boxes contain the x^3-term and the constant term.
What next? Well, the corner box has area 2x^3 and height 2x, so its width must be x^2. And so the area of the other box in the first column must be (x^2)(-3).
How about the box that is catty-corner from the last one? We know that its area, together with -3x^2, makes 7x^2. Therefore, its area must be 10x^2.
We can continue reasoning in this way to work out the rest of the diagram:
Voila! The width of the box is x^2 + 5x – 1, so this must be the quotient you get when you divide 2x^3 + 7x^2 – 17x + 3 by 2x – 3.
Reflection: in the first go-round, we used “length time width gives area;” in the second go-round, we used “area divided by length gives width.”
Now — all of that seems crystal-clear and wonderful. Students can master this with a bit of practice. Why mess with a good thing? Some people apparently want to abstract the above diagram *even further* and produce this following picture with “just numbers.”
This is, in fact, the synthetic division algorithm at work. Indeed, if you compare this diagram to the one at the top of the post, you’ll see that the 6 intermediate values are the same (except for their signs), and the 3 final values on the top row are also the same.
So what are we to do with all of this? My personal opinion: stick with the box diagram, showing the powers of x explicitly. The box diagram works when the divisor is linear, and it also works when it isn’t. And it’s perfectly clear at all stages what is going on. Students can clearly see that the process of division is merely multiplying in reverse.
“Synthetic division,” however, requires a linear divisor, requires somewhat strange manipulations before and after you start it, and is generally *not transparent.* Its sole advantage is speed.
As for me, I’ll take clear understanding over speed 7 days of the week.