Also 4 is not an upper bound for the positive zeros

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Unformatted text preview: or f ). Also, 3 is not an upper bound for the positive zeros (roots). 3 Trying 4: 2 Coeff x − 11x − 22 x −6 of 3 3 − 11 − 22 − 6 12 3 4 − 72 1 − 18 − 78 4 Thus, q 2 ( 4 ) = − 78 ≠ 0 ⇒ x − 4 is not a factor of q 2 (nor f ) and 4 is not a zero (root) of q 2 (nor f ). Also, 4 is not an upper bound for the positive zeros (roots). 3 2 Coeff x − 11x − 22 x −6 of 3 3 − 11 − 22 −6 Trying 6: 18 3 42 120 7 20 6 114 Thus, q 2 ( 6 ) = 114 ≠ 0 ⇒ x − 6 is not a factor of q 2 (nor f ) and 6 is not a zero (root) of q 2 (nor f ). However, 6 is an upper bound for the positive zeros (roots) since all the numbers are positive in the third row of the synthetic division. 3 Trying − 1 : 3 2 Coeff of 3x − 11x − 22 x − 6 3 − 11 − 22 − 6 −1 3 4 − 18 1 3 6 − 12 − 0 1 1 1 Thus, q 2 − = 0 ⇒ x + is a factor of q 2 (and f ) and − is a zero 3 3 3 (root) of q 2 (and f ). 1 2 Thus, 3 x 3 − 11 x 2 − 22 x − 6 = x + ( 3 x − 12 x − 18 ) = 3 1 2 x + 3 ( x − 4 x − 6 ) = ( 3 x + 1) ( x 2 − 4 x − 6 ) 3 Thus, 3 x 5 − 23 x 4 + 34 x 3 + 38 x 2 − 64 x − 24 = ( x − 2 ) 2 ( 3 x 3 − 11 x 2 − 22 x − 6 ) = ( x − 2 ) 2 ( 3 x + 1 ) ( x 2 − 4 x − 6 ) . Now, we can try to find a factorization for the expression However, it does not factor. x2 − 4x − 6. Thus, 3 x 5 − 23 x 4 + 34 x 3 + 38 x 2 − 64 x − 24 = 0 ⇒ ( x − 2 ) 2 ( 3 x + 1) ( x 2 − 4 x − 6 ) = 0 ⇒ x = 2 , x = − 1 , x2 − 4x − 6 = 0 3 Using the Quadratic Formula to solve x 2 − 4 x − 6 = 0 , we have that x= −b ± b 2 − 4ac 2a 4± 40 2 Answer: = 4±2 10 2 4± = 2 = 2± Zeros (Roots): − Factorization: 16 − 4 ( 1 ) ( − 6 ) = 4± 16 + 24 2 10 1 , 2 (multiplicity 2), 2 + 3 10 , 2 − 10 3 x 5 − 23 x 4 + 34 x 3 + 38 x 2 − 64 x − 24 = ( x − 2 ) 2 ( 3 x + 1) ( x 2 − 4 x − 6 ) =...
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