Synthetic Summer Fun

Today, for some summer fun, let’s look at synthetic division a/k/a synthetic substitution. I’ll assume you all know how to do that since it is a pretty common pre-calculus topic and even comes up again in calculus.

Why Does Synthetic Division Work?

An example: consider the polynomial

$P(x)=2{{x}^{4}}-3{{x}^{3}}-11{{x}^{2}}+14x-1$.

This can be written in nested form like this

$P(x)=((((2x-3)x-11)x+14)x-1)$

To evaluate this last expression at, say x = 2, we do the arithmetic as follows:

1.   2 x 2 – 3 = 1
2.   2 x 1 – 11 = –9
3.   2 x (–9) + 14 = –4
4.   2 x (–4) – 1 = – 9 = f(2)

Notice that this requires only multiplication and addition or subtraction, no raising to powers. More to the point, this is the same arithmetic, in the same order when you do the evaluation by synthetic division, and the work is a little easier to keep track of.

$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$

Synthetic division has another advantage: the other numbers in the second row are the coefficients of a quotient polynomial, a polynomial of one less degree that the original. So,

$\displaystyle \frac{P(x)}{x-2}=2{{x}^{3}}+{{x}^{2}}-9x-4+\frac{-9}{x-2}$

The Remainder Theorem and the Factor Theorem

In general, a polynomial of degree n, divided by a linear factor (x – a) gives a polynomial Q(x) of degree n – 1 and a remainder R

$\displaystyle \frac{P(x)}{x-a}=Q(x)+\frac{R}{x-a}$

Or

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

From here it is easy to see that $P(a)=R$. This is called the remainder theorem. It has a corollary called the factor theorem: If R = 0, then (x – a) is a factor of P(x).

Calculus

But wait there is more: differentiating the equation above using the product rule gives

${P}'(x)=Q(x)(1)+Q(x)(x-a)+0$ and substituting x = a  gives

${P}'(a)=Q(a)$. The value of the quotient polynomial at a is the derivative of the original polynomial at a.

Of course, we could also rewrite the same equation as $\displaystyle \frac{P(x)-P(a)}{x-a}=Q(x)$ . Then

$\displaystyle {P}'(a)=\underset{x\to a}{\mathop{\lim }}\,\frac{P(x)-P(a)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,Q(x)=Q(a)$

Taylor Series

But wait, there’s even more.

A polynomial is a Maclaurin series in which all the terms after the nth term are zero. When you students are first learning how to write a Taylor series, by finding all the derivatives and substituting in the general term, a good exercise is to have them write the Taylor series for a polynomial centered away from the origin. For the example above:

$P(x)=-9-2\left( x-2 \right)+19{{\left( x-2 \right)}^{2}}+13{{\left( x-2 \right)}^{3}}+2{{\left( x-2 \right)}^{4}}$

Then ask them to expand the expression above and collect term etc. They should get the original polynomial again (and have some great practice expand powers of a binomial).

Can synthetic division help us? Yes, of course.  Here, is the original computation again:

$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$

If we ignore the –9 and divide the quotient numbers by 2 we get

$\begin{matrix} {} & 2 & 1 & -9 & -4 \\ 2) & 2 & 5 & 1 & -2 \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(2) \\ \end{matrix}$

$\begin{matrix} {} & 2 & 5 & 1 \\ 2) & 2 & 9 & 19 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}'}'(2)}{2} \\ \end{matrix}$

And again

$\begin{matrix} {} & 2 & 9 \\ 2) & 2 & 13 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{{P}'}'}'\left( 2 \right)}{3!} \\ \end{matrix}$

One more time

$\begin{matrix} {} & 2 \\ 2) & 2 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}^{(4)}}\left( 2 \right)}{4!} \\ \end{matrix}$

What do you see? Right, the last numbers in each computation, –9, –2, 19, 13, and 2, are the coefficients of the Taylor polynomial!

If you really want to dive this home and have some more summer fun here’s the start of a proof (at least for n = 4). Let

$P(x)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{3}}+{{c}_{1}}x+{{c}_{0}}$ and divide this by a:

$\begin{matrix} {} & {{c}_{4}} & {{c}_{3}} & {{c}_{2}} & {{c}_{1}} & {{c}_{0}} \\ a) & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{a}}a+{{c}_{1}} & {{c}_{4}}{{a}^{4}}+{{c}_{3}}{{a}^{3}}+{{c}_{a}}{{a}^{2}}+{{c}_{1}}a+{{c}_{0}}=P(a) \\ \end{matrix}$

Again

$\begin{matrix} {} & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{2}}a+{{c}_{1}} \\ a) & {{c}_{4}} & 2{{c}_{4}}a+{{c}_{3}} & 3{{c}_{4}}{{a}^{2}}+2{{c}_{3}}a+{{c}_{2}} & 4{{c}_{4}}{{a}^{3}}+3{{c}_{3}}{{a}^{2}}+2{{c}_{2}}a+{{c}_{1}} \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(a) \\ \end{matrix}$

And I’ll leave the rest to you.  Really, why should I have all the fun?

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Extremes without Calculus

There was an interesting question last week on the AP Calculus Community Bulletin Board. A teacher was working with polynomial functions in her pre-calculus class and the class had learned how to find the roots, and how the roots and their multiplicities affected the graph. Her question was whether there is a way to determine the number of extreme values (maximums and minimums) without using calculus.

You might want to work this yourself before reading on.

Before we go into more detail recall two facts about polynomial functions with real coefficients:

• As a corollary to the Fundamental Theorem of Algebra, we know that a polynomial function of degree n has exactly n factors some of which may be the same. Therefore, the polynomial has exactly n roots, again not necessarily different. The multiplicity of a root is the number of times the corresponding factor appears in the factorization.
• Between any two consecutive roots there is exactly one turning point. (You may need some calculus to justify this. If there were more than one turning point between two consecutive roots, there must be at least three of them. Then the derivative would have more zeros than the original polynomial, which of course cannot happen.

One reader suggested a method for dealing with cubics which I think generalizes to any polynomial. The method is this: Subtract the constant term from the polynomial. This translates the graph so that it will contain the origin. The translated graph is congruent to the original and therefore will have the same number of extreme values.

Red Graph: $y={{x}^{3}}-4{{x}^{2}}+3x+4$
Blue Graph: $y={{x}^{3}}-4{{x}^{2}}+3x$

Find the Real roots of the translated polynomial.  Then the number of turning points for both polynomials can be found this way (assuming, for the moment, that all the roots are Real numbers):

1. For each distinct root, count 1
2. For each root of even multiplicity, count 1
3. Each root of odd multiplicity greater than one, ( 3, 5, 7, …), count 0

The number of turning points will be one less than the total.

Example $P(x)=x{{\left( x-1 \right)}^{4}}{{\left( x-2 \right)}^{3}}{{\left( x-3 \right)}^{8}}$, the count is 4 + 2 + 0 = 6. Turning points = 5

The reasoning goes something like this (for a polynomial with Real coefficients and constant term of zero):

1. A linear polynomial has one root and no turning points.
2. Each time you multiply by a different linear factor you add one root and one turning point.
3. Each time you multiply by several of the same factor you already used to end with an even power for that factor, you add no roots, but you add one turning point (on the x-axis at the root).
4. Each time you multiply by several of the same factor you already used to end with an odd power for that factor, you add no roots, and no turning points, since the graph will cross the axis at such a value.
5. Then the count scheme above will then give the number of turning points.

Another approach is this: consider a polynomial with n distinct roots and (n – 1) turning points spread out along the number line. Now move two adjacent roots together (resulting in a root of multiplicity 2).  The turning point between them moves onto the x-axis. You have lost a root but not a turning point; thus, in the count item 1 decreases by one, so to compensate you increase item 2 by one to get the same total. Now move a third adjacent root to the place of the other two to get a root of multiplicity 3: this time you lose another root, and a turning point so decreasing item 1 will keep the number of turning points given by the count correct.

Did I miss anything? Yes. I did not consider translated polynomials with Complex roots. That’s because I have not (yet) figured that out.  I’m pretty sure that a unfactorable quadratic factor (i.e.  one with Complex conjugate roots) does not add any turning points to the graph, but I haven’t quite convinced myself.  Any suggestions?