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Polynomial and Synthetic Division Pre-Calculus Mrs.Volynskaya Polynomial Division Polynomial Division is very similar to long division. Example: 3x 3 5 x 2 10x 3 3x 1 Polynomial Division x 2x 2 4 7 3x 1 3x 1 3x3 5x 2 10x 3 3x x 3 2 Subtract!! 6x 10 x 2 6x 2x 12 x 3 12 x 4 2 7 Subtract!! Subtract!! Polynomial Division Example: 2 x 9 x 15 2x 5 3 2 Notice that there is no x term. However, we need to include it when we divide. Polynomial Division x 2 2x 5 10 2x 5 2 x 5 2 x3 9 x 2 0 x 15 2 x 5x 3 2 4x 0 x 2 4 x 10x 10 x 15 10 x 25 2 10 Try This Example: x 5 x 10x 9 x 34 x2 4 3 2 Answer: x 3 3x 2 4 x 17 What does it mean if a number divides evenly into another?? Now let’s look at another method to divide… Why??? Sometimes it is easier… Synthetic Division Synthetic Division is a ‘shortcut’ for polynomial division that only works when dividing by a linear factor (x + b). It involves the coefficients of the dividend, and the zero of the divisor. (SUSPENSE IS BUILDING) Example Divide: Step 1: x 2 5x 6 x 1 Write the coefficients of the dividend in a upside-down division symbol. 1 5 6 Example Step 2: x 2 5x 6 x 1 Take the zero of the divisor, and write it on the left. The divisor is x – 1, so the zero is 1. 1 1 5 6 Example Step 3: x 2 5x 6 x 1 Carry down the first coefficient. 1 1 1 5 6 Example Step 4: x 2 5x 6 x 1 Multiply the zero by this number. Write the product under the next coefficient. 1 1 1 5 1 6 Example Step 5: x 2 5x 6 x 1 Add. 1 1 5 1 1 6 6 Example x 2 5x 6 x 1 Step etc.: Repeat as necessary 1 1 1 5 1 6 6 6 12 Example The numbers at the bottom represent the coefficients of the answer. The new polynomial will be one degree less than the original. x 2 5x 6 x 1 1 1 5 6 1 6 12 x6 x 1 1 6 12 Synthetic Division The pattern for synthetic division of a cubic polynomial is summarized as follows. (The pattern for higher-degree polynomials is similar.) Synthetic Division This algorithm for synthetic division works only for divisors of the form x – k. Remember that x + k = x – (–k). Using Synthetic Division Use synthetic division to divide x4 – 10x2 – 2x + 4 by x + 3. Solution: You should set up the array as follows. Note that a zero is included for the missing x3-term in the dividend. Example – Solution cont’d Then, use the synthetic division pattern by adding terms in columns and multiplying the results by –3. So, you have . Try These Examples: + x3 – 11x2 – 5x + 30) (x – 2) (x4 – 1) (x + 1) [Don’t forget to include the missing terms!] (x4 Answers: + 3x2 – 5x – 15 x3 – x2 + x – 1 x3 Application of Long Division To begin, suppose you are given the graph of f (x) = 6x3 – 19x2 + 16x – 4. Long Division of Polynomials Notice that a zero of f occurs at x = 2. Because x = 2 is a zero of f, you know that (x – 2) is a factor of f (x). This means that there exists a second-degree polynomial q (x) such that f (x) = (x – 2) q(x). To find q(x), you can use long division. Example - Long Division of Polynomials Divide 6x3 – 19x2 + 16x – 4 by x – 2, and use the result to factor the polynomial completely. Example 1 – Solution Think Think Think Multiply: 6x2(x – 2). Subtract. Multiply: –7x(x – 2). Subtract. Multiply: 2(x – 2). Subtract. Example – Solution From this division, you can conclude that 6x3 – 19x2 + 16x – 4 = (x – 2)(6x2 – 7x + 2) and by factoring the quadratic 6x2 – 7x + 2, you have 6x3 – 19x2 + 16x – 4 = (x – 2)(2x – 1)(3x – 2). cont’d Long Division of Polynomials 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). Example Using the Remainder Theorem Use the Remainder Theorem to evaluate the following function at x = –2. f (x) = 3x3 + 8x2 + 5x – 7 Solution: Using synthetic division, you obtain the following. Example – Solution cont’d Because the remainder is r = –9, you can conclude that r = f(k) f (–2) = –9. This means that (–2, –9) is a point on the graph of f. You can check this by substituting x = –2 in the original function. Check: f (–2) = 3(–2)3 + 8(–2)2 + 5(–2) – 7 = 3(–8) + 8(4) – 10 – 7 = –9 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. Example – Factoring a Polynomial: Repeated Division Show that (x – 2) and (x + 3) are factors of f (x) = 2x4 + 7x3 – 4x2 – 27x – 18. Then find the remaining factors of f (x). Solution: Using synthetic division with the factor (x – 2), you obtain the following. 0 remainder, so f(2) = 0 and (x – 2) is a factor. 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 2x2 + 5x + 3 = (2x + 3)(x + 1) the complete factorization of f (x) is f (x) = (x – 2)(x + 3)(2x + 3)(x + 1). 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 x-intercept of the graph.