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Last modified: January 26, 2018

Polynomial Division ------------------- Example 1. (x3 + 4x2 + 3x + 3)/(x2 + x + 2) x + 3 ----------------- x2 + x + 2 |x3 + 4x2 + 3x + 3 x3 + x2 + 2x ----------------- 3x2 + x + 3 3x2 + 3x + 6 ------------- - 2x - 3 x + 3 + (-2x - 3) --------- x2 + x + 2 Example 2. (x3 - 1)/(x + 2) x2 - 2x + 4 ------------------ x + 2 |x3 + 0x2 + 0x - 1 x3 + 2x2 -------------- - 2x2 + 0x - 2x2 - 4x -------- 4x - 1 4x + 8 ------- -9 x2 - 2x + 4 + -9/(x + 2) If the remainder is 0 then the denominator is a factor of the numerator. Synthetic Division ------------------ Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor. anxn + an-1xn-1 + ... a0 ---------------------- (bx ± c) Set (bx ± c) = 0 so x = ±c/b = d d |an an-1 an-2 .............. a0 ------------------------------------ dan dr ds an an-1 an-2 + + dan dr = = r s Take the first number inside, representing the leading coefficient, and carry it down, unchanged, to below the division symbol. Multiply this carry-down value by d, and carry the result up into the next column. Add down the column. Multiply the result by d, and carry result into next column. Add down the column and so on. The last carry-down value is the remainder. Ex. 4x3 + 2x2 - 12 divided by x - 3 3 | 4 2 0 -12 ----------- 12 42 126 4 14 42 114 => 4x2 + 14x + 42 + 114/(x - 3) Again, if the divisor is a factor of the equation, the remainder will be 0. The resulting quotient is referred to as the depressed equation. The depressed equation can be used to find the remaining roots. In the case of a cubic the depressed equation will be a quadratic. In thecase of a quartic, the depressed equation will be a cubic and the above procedure will have to be repeated. Rational Roots Test ------------------- The Rational Roots (or Rational Zeroes) Test is a handy way of obtaining a list of useful first guesses when you are trying to find the zeroes (roots) of a polynomial. - The coefficients have to be integers - This only works if the equation has rational roots (1, 2, 3/2 etc). If not the cubic or quartic formula has to be used. - The test does not give you the zeroes. It does not say what the zeroes definitely will be. It only gives you a list of relatively easy and "nice" numbers to try in the polynomial. Most of these possible zeroes will turn out not actually to be zeroes! anxn + an-1xn-1 + ... a0 Find +/- factors of a0 = p. Find + factors of leading coefficient, an = q. Form all combinations of p/q. Not all results will be roots. Need to sub into original equation to figure out which ones are valid. Ex. 3x3 - 5x2 + 5x - 2 = 0 p = +/-1, +/-2 q = 1, 3 p/q = ±1/1, ±1/3, ±2/1, ±2/3, By trial and error we find that only +2/3 works. Descartes' Rule of Signs ------------------------ Tells you the types of roots you can expect in a polynomial. f(x) = 4x7 + 3x6 + x5 + 2x4 - x3 + 9x2 + 9x + 1 The sign changes 2 times. To get the complete list of positive roots it is necessary to count down by 2's. Thus, in this case, there are either 2 or 0. f(-x) = -4x7 + 3x6 - x5 + 2x4 + x3 + 9x2 - 9x + 1 The sign changes 5 times. Thus, there are either 5, 3 or 1 negative roots. If the sum of the positive and negative roots doesn't equal the order of the equation, then this implies there is also the possibility of complex roots (which must appear in pairs). Thus, + real -real complex total ------ ----- ------- ----- 2 5 0 7 2 3 2 7 2 1 4 7 0 5 2 7 0 3 4 7 0 1 6 7 Example 1. x3 - 8 = 0 has 1 positive root and 0 negative roots. Therefore, it must have 2 complex roots. Positive root: x = 3√8 = 2 2 | 1 0 0 -8 ----------- 2 4 8 1 2 4 0 Depressed equation = x2 + 2x + 4 = 0 with roots -1 +/- √-3 => -1 +/- i√3