1 / 15

Polynomial and Synthetic Division

Polynomial and Synthetic Division. Section 3.3 College Algebra, MATH 171 Mr. Keltner. Graphically, now Algebraically. So far in this chapter we have dealt with polynomial functions graphically. That is, we’ve been mainly interested in how they look .

ferris-wheeler
Download Presentation

Polynomial and Synthetic Division

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Polynomial and Synthetic Division Section 3.3 College Algebra, MATH 171 Mr. Keltner

  2. Graphically, now Algebraically So far in this chapter we have dealt with polynomial functions graphically. That is, we’ve been mainly interested in how they look. Most of our work to follow will deal with factoring polynomials and finding zeros. To factor, we must know how to divide polynomials. We will investigate several different strategies to help us do this.

  3. Simpler Example To refresh us on how to use long division, try the problem 4395 ÷ 18. Think of different ways of expressing your answer, if you encounter a remainder. How could we rewrite the same relationship using multiplication instead?

  4. Division Algorithm, pg. 290 What we have done in our process of long division is found real numbers such that: F(X)÷D(X) = Q(X) + R(X)/D(X) When applying this same principle to long division of polynomials, we are working to find polynomials that yield the same relationships. It would be the same as showing that the two expressions below are equivalent:

  5. Before you get started: Before working with the Division Algorithm, make sure a couple things are in order: Write the dividend and divisor in descending order according to their degrees. Insert placeholders with zero coefficients for missing powers of the variable. This means to rewrite something like x3-7x+1 as x3 +0x2 - 7x + 1.

  6. Same Principle, Different Look Example 1 • Relate this to the long division you are used to seeing from a couple years ago. • Make sure, if a degree is not included in an expression that you mention it by writing 0xn in its place. Using Polynomial Long Division

  7. Checking Your Answer To make sure that you multiplied correctly, we can graph the original expression for y1 and your simplified answer for y2 on a graphing calculator. If you only see one graph, this means the two lines coincide, and that the two expressions are equivalent. If you see more than one line, you did not simplify the expression correctly or you may have left out a term on your calculator. You can always use the distributive property, much like the FOIL method, to check if your expressions are equivalent.

  8. Example 2 USING LONG DIVISION WITH A LINEAR DIVISOR • Divide f(x) = x3 – x2 + 4x – 10 by (x + 2).

  9. Synthetic division can be used to divide any polynomial by a divisor of the form (x – k). • Try Example 2 again, but use synthetic division this time. • Divide f(x) = x3 – x2 + 4x – 10 by (x + 2). Synthetic Division If you use synthetic substitution to evaluate Example 2 when x = -2, as shown below, you can see that f(2) equals the remainder when f(x) is divided by (x + 2). Also, the other values below the line match the coefficients of the quotient. For this reason, synthetic division is sometimes called synthetic division.

  10. Remainder Theorem IF A POLYNOMIAL P(x) IS DIVIDED BY (x - c), THEN THE REMAINDER IS R = P(c).

  11. Example 3 • Use synthetic division to divide: • P(x) = x5 + 5x4 – 4x3 + 7x + 3 ÷ (x + 2) • What is the value of P(-2)?

  12. Factor Theorem From our findings with the Remainder Theorem, we notice if we divide P(x) by (x - c) and get a remainder of zero, it corresponds to a value of y-value at c such that P(c) = 0. THE FACTOR THEOREM STATES THAT: c is a zero of the polynomial P(x) if and only if (x – c) is a factor of P(x). [If P(c = 0, then (x - c) is a factor of P(x).] THE FACTOR THEOREM ALLOWS US TO SOLVE A VARIETY OF PROBLEMS.

  13. Example 4 Factor the polynomial f(x) = x3 – 6x2 + 5x + 12, given that (x – 4) is a factor.

  14. Example 5 • One zero of f(x) = x3 + 8x2 + 5x – 14 is f(-2) = 0. Find the other zeros of f(x) and use these to factor f(x) completely.

  15. Check for Understanding Pgs. 295-296: #’s 1, 2, 5-60, multiples of 5

More Related