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Math 135 Synthetic Division Examples Polynomial Long Division Is A Math 135 Synthetic Division Examples Polynomial long division is a tedious process which can be shortened considerably in the special case when the divisor is a linear factor. By the factor theorem, if we divide a polynomial p(x) by the linear polynomial x − c, then p(c) is the remainder and p(c) = 0 if and only if (x − c) is a factor of p(x). This observation suggest that in order to factor a polynomial p(x) of large degree it suffices to look for numbers c such that p(c) = 0. This is equivalent to long division by (x−c), which can be extremely drawn out. The method of synthetic division accomplishes division by a linear factor quickly and is done as follows: x6−64 Example 1. Divide x−2 . Begin as with polynomial long division but write only the coefficients of x6−64 making sure to list all of them. [2] 1 0 0 0 0 0 -64 0 1 Carry the coefficient of the leading term as shown and then multiply by c = 2, add to the next column and repeat. Thus [2] 1 0 0 0 0 0 -64 0 2 4 8 16 32 64 1 2 4 8 16 32 [0] The numbers below the line are the coefficients of a polynomial of one degree lower, i.e. x5 + 2x4 + 4x3 + 8x2 + 16x + 32 and the last number is the remainder. Therefore, x6 − 64 = x5 + 2x4 + 4x3 + 8x2 + 16x + 32 x − 2 Observe that finding roots of any polynomial p(x) is now reduced to finding a number c (that would play the role of 2 above) such that the remainder of the synthetic division, namely p(c), is zero. University of Hawai‘i at Manoa¯ 125 R Spring - 2014 Math 135 Synthetic Division Examples Example 2. Factor and find all roots of x4 − 6x3 − 11x2 + 24x + 28. We begin with small test values c. Test first 1; −1; 2; −2;:::. We have p(1) = 36, but p(−1) = 0 and [-1] 1 -6 -11 24 28 0 -1 7 4 -28 1 -7 -4 28 [0] in other words we have performed the division x4 − 6x3 − 11x2 + 24x + 28 = x3 − 7x2 − 4x + 28 x + 1 Repeating this process again with the newly obtained polynomial we see that −2 works and hence [-2] 1 -7 -4 28 0 -2 18 -28 1 -9 14 [0] in other words we have performed the division x3 − 7x2 − 4x + 28 = x2 − 9x + 14 = (x − 7)(x − 2) x + 2 Therefore, x4 − 6x3 − 11x2 + 24x + 28 = (x − 7)(x − 2) (x + 1)(x + 2) and x4 − 6x3 − 11x2 + 24x + 28 = (x + 1)(x + 2)(x − 7)(x − 2) and the roots are {−2; −1; 2; 7g. University of Hawai‘i at Manoa¯ 126 R Spring - 2014 Math 135 Synthetic Division Worksheet 1. Perform the division. List the quotient and remainder. 3x2−11x+5 (a) x−4 5x5+3x3+1 (b) x+2 9x3+14x−6 (c) 3x−2 2. What is the remainder of the division of p(x) by x − 3 if: (a) p(x) = 3x4 + 3x − 1 (b) p(x) = 7x5 − 500x + 3 (c) p(x) = 4x4 + x 3. Find all roots. (a) x3 − 2x2 − 5x + 6 (b) x4 + 2x3 − 9x2 − 2x + 8 University of Hawai‘i at Manoa¯ 127 R Spring - 2014.
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