Polynomial and Synthetic Division


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1 Polynomial and Synthetic Division
2 Polynomial Division Polynomial Division is very similar to long division. Example: 3x 3 5x 3x 10x 1 3
3 Polynomial Division 3x 1 x 3x 3 3 x 5x 3x x 6x 4 10x 10x 7 3 x 1 3 6x x 1x 3 1x 4 Subtract!! Subtract!! Subtract!! 7
4 Polynomial Division Example: x 3 9x 15 x 5 Notice that there is no x term. However, we need to include it when we divide.
5 Polynomial Division x 5 x x 3 3 9x x 5x x 4x 5 0x 15 0x 4x 10x 10x 10x 10 x
6 Try This Example: x 4 5x 3 10x 9x 34 x Answer: x 3 3x 4x 17
7 Now let s look at another method to divide Why??? Sometimes it is easier
8 Synthetic Division Synthetic Division is a shortcut for polynomial division that only works when dividing by a linear factor (x  a). It involves the coefficients of the dividend, and the zero of the divisor.
9 Example Divide: Step 1: x 5x 6 x 1 Write the coefficients of the dividend in a upsidedown division symbol
10 Example Step : x Take the zero of the divisor, and write it on the left. The divisor is x 1, so the zero is x 6 x
11 Example Step 3: x 5x 6 x 1 Carry down the first coefficient
12 Example Step 4: x 5x 6 x 1 Multiply the zero by this number. Write the product under the next coefficient
13 Example Step 5: Add. x 5x x
14 Example Step etc.: x 5x 6 x 1 Repeat as necessary
15 Example The numbers at the bottom represent the coefficients of the answer. The new polynomial will be one degree less than the original. x 5x 6 x x x 1
16 Synthetic Division The pattern for synthetic division of a cubic polynomial is summarized as follows. (The pattern for higherdegree polynomials is similar.)
17 Synthetic Division This algorithm for synthetic division works only for divisors of the form x a. Remember that x + a = x ( a).
18 Using Synthetic Division Use synthetic division to divide x 4 10x x + 4 by x + 3. Solution: You should set up the array as follows. Note that a zero is included for the missing x 3 term in the dividend.
19 Example Solution cont d Then, use the synthetic division pattern by adding terms in columns and multiplying the results by 3. So, you have.
20 Try These Examples: (x 4 + x 3 11x 5x + 30) (x ) (x 4 1) (x + 1) [Don t forget to include the missing terms!] Answers: x 3 + 3x 5x 15 x 3 x + x 1
21 Application of Long Division To begin, suppose you are given the graph of f (x) = 6x 3 19x + 16x 4.
22 Long Division of Polynomials Notice that a zero of f occurs at x =. Because x = is a zero of f, you know that (x ) is a factor of f (x). This means that there exists a seconddegree polynomial q (x) such that f (x) = (x ) q(x). To find q(x), you can use long division.
23 Example  Long Division of Polynomials Divide 6x 3 19x + 16x 4 by x, and use the result to factor the polynomial completely.
24 Example 1 Solution Think Think Think Multiply: 6x (x ). Subtract. Multiply: 7x(x ). Subtract. Multiply: (x ). Subtract.
25 Example Solution cont d From this division, you can conclude that 6x 3 19x + 16x 4 = (x )(6x 7x + ) and by factoring the quadratic 6x 7x +, you have 6x 3 19x + 16x 4 = (x )(x 1)(3x ).
26 Long Division of Polynomials
27 The Remainder and Factor Theorems
28 The Remainder and Factor Theorems The remainder obtained in the synthetic division process has an important interpretation, as described in the Remainder Theorem. The Remainder Theorem tells you that synthetic division can be used to evaluate a polynomial function. That is, to evaluate a polynomial function f (x) when x = k, divide f (x) by x k. The remainder will be f (k).
29 Example Using the Remainder Theorem Use the Remainder Theorem to evaluate the following function at x =. f (x) = 3x 3 + 8x + 5x 7 Solution: Using synthetic division, you obtain the following.
30 Example Solution cont d Because the remainder is r = 9, you can conclude that f ( ) = 9. r = f(k) This means that (, 9) is a point on the graph of f. You can check this by substituting x = in the original function. Check: f ( ) = 3( ) 3 + 8( ) + 5( ) 7 = 3( 8) + 8(4) 10 7 = 9
31 The Remainder and Factor Theorems Another important theorem is the Factor Theorem, stated below. This theorem states that you can test to see whether a polynomial has (x k) as a factor by evaluating the polynomial at x = k. If the result is 0, (x k) is a factor.
32 Example Factoring a Polynomial: Repeated Division Show that (x ) and (x + 3) are factors of f (x) = x 4 + 7x 3 4x 7x 18. Then find the remaining factors of f (x). Solution: Using synthetic division with the factor (x ), you obtain the following. 0 remainder, so f() = 0 and (x ) is a factor.
33 Example Solution cont d Take the result of this division and perform synthetic division again using the factor (x + 3). 0 remainder, so f( 3) = 0 and (x + 3) is a factor. Because the resulting quadratic expression factors as x + 5x + 3 = (x + 3)(x + 1) the complete factorization of f (x) is f (x) = (x )(x + 3)(x + 3)(x + 1).
34 The Remainder and Factor Theorems For instance, if you find that x k divides evenly into f (x) (with no remainder), try sketching the graph of f. You should find that (k, 0) is an xintercept of the graph.
35 Can you Use long division to divide polynomials by other polynomials. Use synthetic division to divide polynomials by binomials of the form (x k). Use the Remainder Theorem and the Factor Theorem.
36 Questions???
37 Homework 9.3 Page # 7, 9, 13, 15, 19, 3, 5, 46, 49, 57 Quiz next time  quadratics
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