“I’m guessing that all of the real-number zeros are in the interval [𝒂, 𝒃]” Use Synthetic Division to check your guesses Testing your upper bound guess, b Use Synthetic Division to divide 𝑃(𝑥) ÷ (𝑥 − 𝑏). 𝑏 𝑎𝑛 * 𝑞𝑛−1 ⋯ * ⋯ 𝑎1 * 𝑞0 𝑎0 * 𝑟 Coefficients of 𝑃(𝑥) Synthetic division Coefficients of the Quotient and Remainder If the remainder 𝒓 = 𝟎, then 𝑏 is a zero of 𝑃(𝑥) if we’re still interested in an upper bound, we must guess a higher value of 𝑏 and repeat the testing procedure. But if the remainder 𝒓 ≠ 𝟎, check the signs of all the bottom row results, all of 𝑞𝑛−1 ⋯, 𝑞0 , 𝑟 If they’re all ≥ 0, then YES, , 𝑏 is an upper bound. Any real-number zeros of 𝑃(𝑥) must be ≤ 𝑏. Otherwise, your guess is wrong. The upper bound is higher than what you guessed. Testing your lower bound guess, a Use Synthetic Division to divide 𝑃(𝑥) ÷ (𝑥 − 𝑎). (Note: the 𝑎 name for our upper bound is unrelated to the 𝑎𝑠𝑢𝑏𝑠𝑐𝑟𝑖𝑝𝑡 names of the coefficients of 𝑃(𝑥). An unfortunate naming coincidence.) 𝑎 𝑎𝑛 * 𝑞𝑛−1 ⋯ * ⋯ 𝑎1 * 𝑞0 𝑎0 * 𝑟 Coefficients of 𝑃(𝑥) Synthetic division Coefficients of the Quotient and Remainder If the remainder 𝒓 = 𝟎, then 𝑎 is a zero of 𝑃(𝑥) if we’re still interested in a lower bound, guess a lower value of 𝑎 and retest. But if the remainder 𝒓 ≠ 𝟎, check the signs of all the bottom row results, all of 𝑞𝑛−1 ⋯, 𝑞0 , 𝑟 If the signs on the bottom row, 𝑞𝑛−1 ⋯, 𝑞0 , 𝑟 alternate between positive and negative, +, −, +, −, ⋯ or −, +, −, +, ⋯, then YES, 𝑎 is a lower bound. Any real-number zeros of 𝑃(𝑥) must be ≥ 𝑎. , then YES, , 𝑏 is an upper bound. Any real-number zeros of 𝑃(𝑥) must be ≤ 𝑏. Otherwise, your guess is wrong. The lower bound is lower than what you guessed. Typical Guesses 𝑎 = −10 and 𝑏 = 10, because those are the WINDOW Xmin and Xmax values you get from the TI-84 ZOOM 6:ZStandard command. If you use the Rational Zero Theorem, there is a largest and smallest candidate for rational number zeros, the ± of the constant term. Those can be good candidates for the bounded zeros analysis. Document1 2/6/2016 10:59 PM - D.R.S.