Sunday, October 24, 2010

Motivating Synthetic Division and Associated Theorems and Rules

So this is what I use to help students practice the various theorems and rules typically associated with synthetic division after we've covered the rules in class. The set-up is "Mr. H found the following work shown by a student on scratch paper during a test. The student did not write the original question, but somehow Mr. H knew what the student was trying to do. Mr. H also noticed that the student could have saved a lot of time during the test had the student used the theorems and rules learned in class. What was the student trying to do? How could the student accomplish the same thing but faster and why? Comment on each step."

The theorems and rules I'm asking students to apply are:
  • Intermediate Value Theorem
  • Remainder Theorem
  • Factor Theorem
  • The Rational Zero Test
  • Descarte's Rule of Signs
  • Upper Bound and Lower Bound Rules

We use Pre-Calculus with Limits: A Graphing Approach by Larson, Hostetler, and Edwards.

Some comments that I'm looking for:
  1. p/q is work for the Rational Zero Test. It is used to find all the possible rational zeros.
  2. f(x)=2x3+5x2-37x-60 has 1 variation in sign (sign change in the coefficients) so there's 1 positive real zero by Descarte's Rule of Signs. (Work is not shown)
  3. f(-x)=-2x3+5x2+37x-60 has 2 variations in sign (sign change in the coefficients) so there's 2 or 0 negative real zeros by Descarte's Rule of Signs. (Work is not shown)
  4. 5 is an upper bound because the last row is either positive or zeros and 5 is positive. This means no real zeros above 5.
  5. 6 should be skipped since 5 is an upper bound. However, this may still be useful if we're graphing the function. The remainder of 330 means f(6)=330. So we have as point (6,330).
  6. 3 has a remainder of -72. Since f(3)=-72 and f(5)=180, and since f(x) is a polynomial (continuous everywhere), there must be a zero in the interval [3,5] by Intermediate Value Theorem. 4 is a good number to try.
  7. 2 is unnecessary since we know there is only 1 positive zero and it is between 3 and 5.
  8. 4 happens to be a zero. The quotient is 2x2+13x+15. Student should attempt to factor this quadratic to save time.
  9. 1 is unnecessary since there is only 1 positive real zero by Descarte's Rule of Signs and it's been found that 4 is that zero.
  10. -6 is a lower bound since the last row is alternately positive and negative and -6 is negative. This means no real zeroes below -6. The original f(x) should not have been used here. The quotient of 2x2+13x+15 should have been used instead, if factoring is not immediately obvious.
  11. -10 should not have been tried if all we are doing is factoring. We already know -6 is a lower bound.
  12. -5 is a zero.
  13. 4 is used here to find out the last factor. We found 4 was a zero from an earlier step.
  14. The polynomial f(x)=2x3+5x2-37x-60 was factored to f(x)=(x-4)(x+5)(2x+3)

It seems to me that, for many students, time is better spent practicing how to consistently and accurately divide using synthetic division. Here's some questions to math teachers:
  1. Do you teach these theorems and rules?
  2. If you don't cover them, why not?
  3. If you do, how do you cover them (how much time/depth)?
  4. How do you teach/motivate learning these theorems and rules?

I'd like to hear your thoughts.

1 comment: