Lesson 3 - by an even number from this maximum. So, f ( 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. Vietas 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 rst 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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