Lesson 3

# Lesson 3 - by an even number from this maximum So f x = 3 x...

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Finding Roots of Polynomials Let’s start with a quadratic: p q x 2 + r s x + t u = 0 with p,q,r,s,t,u all integers Then ﬁnd a common denominator and a common factor and divide them out: commonfactor commondenominator ( ax 2 + x + c ) = 0 So we call f ( x ) = ax 2 + bx + c This gives a new problem, ﬁnd roots of f ( x ) = 0 Now we use Descartes Rule of Signs : The number of sign changes in the coeﬃcients of f ( x ) (meaning we list the coeﬃcients from ﬁrst to last and count how many times they change from positive to negative) tells us the maximum number of positive roots the polynomial has, and the number of sign changes in the coeﬃcients of f ( - x ) gives us the maximum number of negative roots the polynomial has. Furthermore, the actual number of positive or negative roots will diﬀer
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Unformatted text preview: by an even number from this maximum. So, f ( x ) = 3 x 2 + 2 x + 5 has no positive roots and has 2 or 0 negative roots. Vieta’s Formula says that real roots of f ( x ) must have a numerator which is a divisor of the last term c and a denominator which is a divisor of the ﬁrst term a . Last but not least, all of the roots must add up to-b/a . This week, we will practice using these rules where the maximum number of roots is the number of roots and all roots happen once Test for a Root f ( x ) has a root a when f ( x ) = ( x-a ) g ( x ) or f ( a ) = 0 Then use long division to remove the root from the polynomial....
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## This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Fall '10 term at Carleton.

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