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10TH EDITION COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER 3.2 - 1 3.2 Synthetic Division Synthetic Division Evaluating Polynomial Functions Using the Remainder Theorem Testing Potential Zeros 3.2 - 2 Division Algorithm Let (x) and g(x) be polynomials with g(x) of lower degree than (x) and g(x) of degree one or more. There exists unique polynomials q(x) and r(x) such that f x g x q x r x , where either r(x) = 0 or the degree of r(x) is less than the degree of g(x). 3.2 - 3 Synthetic Division Synthetic division is a shortcut method of performing long division with polynomials. It is used only when a polynomial is divided by a first-degree binomial of the form x – k, where the coefficient of x is 1. 3.2 - 4 Synthetic Division 3x 10x 40 3 2 x 4 3 x 2 x 0 x 150 3 x 3 12 x 2 10 x 2 0 x 2 10 x 40 x 40x 150 40x 160 10 2 3.2 - 5 Synthetic Division 3 10 40 Here the division process is simplified 4 3 2 0 150 by omitting the 3 12 variables and writing 10 0 only coefficients, with 10 40 0 used to represent the coefficient of any 40 150 missing terms. 40 160 10 3.2 - 6 Synthetic Division The numbers in color that are repetitions of the numbers directly above them can be omitted as shown here. 3 4 3 10 40 2 0 150 12 10 0 40 40 150 160 10 3.2 - 7 Synthetic Division 4 3 2 0 150 12 40 160 40 10 3 10 The entire problem can now be condensed vertically, and the top row is obtained by subtracting –12, –40, and –160 from the corresponding terms above them. 3.2 - 8 Synthetic Division Additive inverse Quotient 2 0 150 12 40 160 40 10 3 10 43 10 3 x 10 x 40 x4 2 Signs changed Remainder With synthetic division it is helpful to change the sign of the divisor, so the –4 at the left is changed to 4, which also changes the sign of the numbers in the second row. To compensate for this change, subtraction is changed to addition. 3.2 - 9 Caution To avoid errors, use 0 as the coefficient for any missing terms, including a missing constant, when setting up the division. 3.2 - 10 USING SYNTHETIC DIVISION Example 1 Use synthetic division to divide 5 x 6 x 28 x 2 . x2 Solution Express x + 2 in the form x – k by writing it as x – (–2). Use this and the coefficients of the polynomial to obtain x + 2 leads to –2 2 5 3 2 6 28 2. Coefficients 3.2 - 11 Example 1 USING SYNTHETIC DIVISION Use synthetic division to divide 5 x 6 x 28 x 2 . x2 Solution Bring down the 5, and multiply: –2(5) = –10 2 5 6 28 2 10 5 3 2 3.2 - 12 Example 1 USING SYNTHETIC DIVISION Use synthetic division to divide 5 x 6 x 28 x 2 . x2 Solution Add –6 and –10 to obtain –16. Multiply –2(–16) = 32. 3 2 5 2 6 28 10 32 5 16 2 3.2 - 13 Example 1 USING SYNTHETIC DIVISION Use synthetic division to divide 5 x 6 x 28 x 2 . x2 Solution Add –28 and 32 to obtain 4. Finally, –2(4) = – 8. 3 2 5 2 6 28 10 32 5 16 4 2 8 Add columns. Watch your signs. 3.2 - 14 Example 1 USING SYNTHETIC DIVISION Use synthetic division to divide 5 x 6 x 28 x 2 . x2 Solution Add –2 and –8 to obtain –10. 3 2 2 5 6 28 2 10 32 8 5 16 4 10 Remainder Quotient 3.2 - 15 Example 1 USING SYNTHETIC DIVISION Use synthetic division to divide 5 x 6 x 28 x 2 . x2 Since the divisor x – k has degree 1, the degree of the quotient will always be written one less than the degree of the polynomial to be divided. Thus, 3 2 5 x 6 x 28 x 2 10 2 5 x 16 x 4 . x2 x2 Remember to 3 2 add remainder divisor . 3.2 - 16 Special Case of the Division Algorithm For any polynomial (x) and any complex number k, there exists a unique polynomial q(x) and number r such that f ( x ) ( x k )q( x ) r . 3.2 - 17 For Example In the synthetic division in Example 1, 5 x 3 6 x 2 28 x 2 ( x 2)(5 x 2 16 x 4) ( 10). f (x) (x k ) q( x ) r Here g(x) is the first-degree polynomial x – k. 3.2 - 18 Remainder Theorem If the polynomial (x) is divided by x – k, the remainder is equal to (k). 3.2 - 19 Remainder Theorem In Example 1, when (x) = 5x3 – 6x2 – 28x – 2 was divided by x + 2 or x –(–2), the remainder was –10. Substituting –2 for x in (x) gives f ( 2) 5( 2)3 6( 2)2 28( 2) 2 40 24 56 2 10 Use parentheses around substituted values to avoid errors. 3.2 - 20 Remainder Theorem A simpler way to find the value of a polynomial is often by using synthetic division. By the remainder theorem, instead of replacing x by –2 to find (–2), divide (x) by x + 2 using synthetic division as in Example 1. Then (–2) is the remainder, –10. 2 5 6 28 2 10 32 8 (–2) 5 16 4 10 3.2 - 21 Example 2 APPLYING THE REMAINDER THEOREM Let (x) = –x4 + 3x2 – 4x – 5. Use the remainder theorem to find (–3). Solution Use synthetic division with k = –3. 3 1 0 3 4 5 3 9 18 42 1 3 6 14 47 Remainder By this result, (–3) = –47. 3.2 - 22 Testing Potential Zeros A zero of a polynomial function is a number k such that (k) = 0. The real number zeros are the x-intercepts of the graph of the function. The remainder theorem gives a quick way to decide if a number k is a zero of the polynomial function defined by (x). Use synthetic division to find (k); if the remainder is 0, then (k) = 0 and k is a zero of (x). A zero of (x) is called a root or solution of the equation (x) = 0. 3.2 - 23 Example 3 DECIDING WHETHER A NUMBER IS A ZERO Decide whether the given number k is a zero of (x). 3 2 f ( x ) x 4 x 9 x 6; k 1 a. Solution Proposed zero 11 4 9 6 1 3 6 1 3 6 0 f ( x ) x 3 4x 2 9x 6 Remainder Since the remainder is 0, (1) = 0, and 1 is a zero of the polynomial function defined by (x) = x3 – 4x2 + 9x – 6. An x-intercept of the graph (x) is 1, so the graph includes the point (1, 0). 3.2 - 24 Example 3 DECIDING WHETHER A NUMBER IS A ZERO Decide whether the given number k is a zero of (x). 4 2 f ( x ) x x 3 x 1; k 4 b. Solution Remember to use 0 as coefficient for the missing x3-term in the synthetic division. Proposed zero 4 1 0 1 3 1 4 16 68 284 1 4 17 71 285 Remainder The remainder is not 0, so – 4 is not a zero of (x) = x4 +x2 – 3x + 1. In fact, (– 4) = 285, indicating that (– 4, 285) is on the graph of (x). 3.2 - 25 Example 3 DECIDING WHETHER A NUMBER IS A ZERO Decide whether the given number k is a zero of (x). 4 3 2 c. f ( x ) x 2 x 4 x 2 x 5; k 1 2i Solution Use synthetic division and operations with complex numbers to determine whether 1 + 2i is a zero of (x) = x4 – 2x3 + 4x2 + 2x – 5. 1 2i 1 2 4 1 2i 5 (1 2i )( 1 2i ) 1 4 i 2 5 1 1 2i 1 2 5 1 2i 5 1 2i 0 Remainder 3.2 - 26 Example 3 DECIDING WHETHER A NUMBER IS A ZERO Decide whether the given number k is a zero of (x). 4 3 2 c. f ( x ) x 2 x 4 x 2 x 5; k 1 2i Since the remainder is 0, 1 + 2i is a zero of the given polynomial function. Notice that 1 + 2i is not a real number zero. Therefore, it would not appear as an x-intercept in the graph of (x). 1 2i 1 2 4 1 2i 5 (1 2i )( 1 2i ) 1 4 i 2 5 1 1 2i 1 2 5 1 2i 5 1 2i 0 Remainder 3.2 - 27