343 3.2 Synthetic Division We can illustrate this connection using the mathematical statement 5x3 - 6x2 - 28x - 2 = 1x + 2215x2 - 16x + 42 + 1-102. (111111111)1111+1111* (1)1* (1111111)1111111* (1)1* ƒ1x2 = 1x - k2 # q1x2 + r This form of the division algorithm is used to develop the remainder theorem. Special Case of the Division Algorithm For any polynomial ƒ1x2 and any complex number k, there exists a unique polynomial q1x2 and number r such that the following holds true. ƒ1x2 = 1x −k2q1x2 +r Remainder Theorem Suppose ƒ1x2 is written as ƒ1x2 = 1x - k2q1x2 + r. This equality is true for all complex values of x, so it is true for x = k. ƒ1k2 = 1k - k2q1k2 + r, or ƒ1k2 = r Replace x with k. This proves the following remainder theorem, which gives a new method of evaluating polynomial functions. Remainder Theorem If a polynomial ƒ1x2 is divided by x - k, then the remainder is equal to ƒ1k2. In Example 1, when ƒ1x2 = 5x3 - 6x2 - 28x - 2 was divided by x + 2, or x - 1-22, the remainder was -10. Substitute -2 for x in ƒ1x2. ƒ1-22 = 51-223 - 61-222 - 281-22 - 2 ƒ1-22 = -40 - 24 + 56 - 2 ƒ1-22 = -10 An alternative way to find the value of a polynomial is to use synthetic division. By the remainder theorem, instead of replacing x by -2 to find ƒ1-22, divide ƒ1x2 by x + 2 as in Example 1. Then ƒ1-22 is the remainder, -10. -2)5 -6 -28 -2 -10 32 -8 5 -16 4 -10 ƒ1-22 Use parentheses around substituted values to avoid errors. f1-32 is equal to the remainder when dividing by x + 3. EXAMPLE 2 Applying the RemainderTheorem Let ƒ1x2 = -x4 + 3x2 - 4x - 5. Use the remainder theorem to find ƒ1-32. 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, ƒ1-32 = -47. S Now Try Exercise 37.
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