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5.5 Theorems About Roots of Poly Equations The Rational Root Theorem If p/q is in simplest form and is a rational root of a polynomial equation, then p must be a factor of a0 and q must be a factor of an. This says that any roots must be a factor of the coefficient of the last term over a factor of the coefficient of the first term. Example: Find all possible roots of x3-4x2-2x+8 = 0. We need factors of 8 over factors of 1. Factors of 8: plus or minus 1,2,4,8 Factors of 1: plus or minus 1 The possible factors of the poly are plus or minus 1/1,2/1,4/1,8/1. We can then check these to find actual roots using synthetic division. Conjugate Root Theorem Let a and b be rational numbers and let √b be an irrational number. If a+ √b is a root of a poly then the conjugate a- √b is also a root. Example: If 1+ √3 and - √11 are roots give me two more roots? Answer: 1- √3 and √11. Imaginary Root Theorem If the imaginary number a+bi is a root of a poly, then the conjugate a-bi is also a root. Example: Find two more roots if 3i and -2+i are both roots. Answer: -3i and -2-i. Desartes’ Rule of Signs Let P(x) be a poly with real coefficients written in standard form. • The number of positive real roots of P(x) = 0 is either equal to the number of sign changes between consecutive coefficients of P(x) or is less than that by an even number. • The number of negative real roots is either equal to the number of sign changes between consecutive coefficients of P(-x) or is less than that by an even number. • WHAT? Example: Tell about the nature of the roots of x3 – x2 + 1 = 0 There are two positive sign changes, so we know that there are either 2 or 0 positive roots. There is only one negative sign change so we know that we only have one negative root. We can use these theorems to rewrite the original poly. Example: Find a 3rd degree poly with roots --1,2-i. We know that (x+1)(x-(2-i))(x-(2+i)) are the factors. Now foil and distribute to find the poly. (x+1)(x-(2-i))(x-(2+i))= foil (x+1)(x2-x(2-i)-x(2+i)+(2-i)(2+i))= simplify (x+1)(x2-2x+xi-2x-xi+4-2i+2i-i2)= simplify (x+1)(x2-4x+5)= distribute x(x2-4x+5)+1(x2-4x+5)= simplify x3-4x2+5x+x2-4x+5= x3-3x2+x+5=0 HW pg 316 9 , 12 , 18, 23, 24, 31, 37 Lets do 37 now