Questions, questions, questions... It seems like synthetic division has struck a cord..and I get responses asking or telling me that

a) No one should teach synthetic division;

b) I didn't show you the way to divide by linear terms like 3x-5 directly, and I had shown someone in my class a few years ago (and they remembered,... "thank you, math Spirits)...

c) I should have shown you how to find the derivative like I showed their calculus class (not sure how far back that was..I've done it a couple of times)

My answer to a) is, "If you don't want to teach synthetic division, don't; but please give me the freedom to make that choice for myself."

I will try to answer b in my next blog (soon...)

That leaves c) so today I will illustrate that it is almost as easy to find the value of the derivative at a given value using synthetic division as it is to find the value of a function.

I will use the simple f(x) = x

^{3}- 3x

^{2}+ 6x - 7 and we wish to evaluate f(2) and f '(2) ...

From the factor-remainder theorem we know that if a polynomial f(x) is divided by x - b, the remainder is f(b)... so we can simply find the value by synthetic division as below.

From this we see that f(2) = 1... but how does that help us find the derivative... Calculus teachers know that f '(x) = 3x

^{2}- 6x + 6 , and nothing like that seems to pop out of what we see above... and if we evaluate that for x=2 we get 6... but where is it in the synthetic division?

Patience,... one more run... now evaluate the quotient of the above problem at 2 again....and

Behold, as Brahmagupta supposedly wrote.. the remainder of this second division is the first derivative...... and I know your heart is thumping in your chest, wondering, wishing.... what if we did it again, would it be,........ could it be?

Oh, Yeah... and now the descent through the derivatives in similar form is probably clear...

SO WHY does that work?

Let's walk through the second part using some pronumeral value x=a instead of a number

Notice that in the place where we expect the remainder, we see that we get the evaluation f(a)... and when we continue down, you can see the derivative terms accumulate as the division works across to finally reveal, that f '(a) does indeed become 3a

^{2}-6a + 6...

I love math.... You just don't have clever things like that pop out in any other discipline...

For those who would like a slightly more technical explanation, her is one from William Rose:

Write P(x) = (x-a)Q(x) + r

By Remainder Theorem, this is:

P(x) = (x-a)Q(x) + P(a)

P'(a) = lim as x--> a of [ P(x) - P(a) ]/(x-a)

= lim as x--> a of [(x-a)Q(x) + P(a) - P(a)]/(x-a)

=lim as x--> a of Q(x)

=Q(a)

So the theorem says that the quotient after dividing P by a, evaluated at a, is the derivative of P at a. This seems totally transparent when you look at P(x) = (x-a)Q(x) + P(a) in that form. Q(a) is the rate at which P is changing near x =a.