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7.4-7.6

7.4-7.6. Writing Equations from Roots. (Remember that the conjugate is also a root). Double root at -2/3, root at 0, and root at 2i. So -2i is also a root. (x-0)(x + 2/3) (x + 2/3) (x – 2i) (x + 2i). (x)(3x + 2) (3x + 2) (x – 2i) (x + 2i). (x)(9x 2 + 12x + 4)(x – 4i 2 )(x 2 + 4) .

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7.4-7.6

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  1. 7.4-7.6

  2. Writing Equations from Roots (Remember that the conjugate is also a root) Double root at -2/3, root at 0, and root at 2i So -2i is also a root (x-0)(x + 2/3) (x + 2/3) (x – 2i) (x + 2i) (x)(3x + 2) (3x + 2) (x – 2i) (x + 2i) (x)(9x2 + 12x + 4)(x – 4i2)(x2 + 4) (x)(9x4 +12x3+ + 40x2 + 48x + 16) Y=(9x5 + 12x4 + 40x3 + 48 x2 + 16x)

  3. 3+4i and 2 are roots (x – 2) (x – (3 + 4i)(x – (3 – 4i)) (x – 2) (x – 3 - 4i)(x – 3 + 4i) +16 (x – 2) (x2 -3x + 4ix – 3x + 9 – 12i – 4ix + 12i – 16i2 ) (x – 2) (x2 -6x + 25) (x3-6x2 + 25x – 2x2 + 12x- 50) Y = x3 - 8x2 + 37x - 50

  4. Possible Rational Zeros Possible rational zeros: Factors of Constant Factors of lead. coeff Ex: Find possible rational zeros of: 3x4 + ……… + 14 1,2,7,14 1, 3 + { 1, 2, 7, 14, 1/3, 2/3, 7/3, 14/3}

  5. Division 150 75 3 25 5 5 X6 + ………. 12 = (x + ~)(x5 + ………) (x+~)(x4 + ……) (x+~)(x3 + ……) (x + ~)(x2 + ~~~~~) Quad. formula If it’s x6 then find 4 zeros and divide 4 times until x2 If it’s x3 then find 1 zero and divide 1 time until x2

  6. Finding possible rational zeros, zeros and factors 1,2,5,10,2550 1 +1, + 2 +,5, + 10, + 50,+ 25 Y = x3 -8x2 + 37x - 50 Graph to find root(s)… Root is at 2 so 1 real, 2 imaginary 1 - 8 37 -50 0 2 -12 50 1 -6 25 0 Factors (x – 2)(X2 -6x + 25) This means that f(2) = 0 X2 -6x + 25 Not factorable so use Q.formula…… 3 + 4i and 3 – 4i Zeros: 2, 3 + 4i, 3 – 4i

  7. G(x) = 48x4 - 52x3 + 13x - 3 Possible Rational : 1, 3 Zeros 1,2,3,4,6,8,12,16, 24,48 +{1, 1/2, 1/3, 1/4, 1/6, 1/8, 1/12, 1/6,1/24, 1/48, 3, 3/2, ¾, 3/8, 3/16} Graph and zeros are at -1/2, ½, ¾, 1/3 -1/2 48 - 52 0 13 -3 0 -24 38 -19 3 48 -76 38 -6 0 This is 48x3 so keep going ½ 0 24 -26 6 48 -52 12 0 (x + ½)(x – ½)4(4x – 3)(3x – 1) (2x + 1)(2x – 1)(4x-3)(3x – 1) 48X2 -52x + 12  4(12x2 -13x + 3) 4(4x – 3)(3x – 1)  1/3 and ¾ or Q.F

  8. Using synthetic to find f(~)Just use ~ in synthetic substitution Given a factor (x + #), find other factors Just use -~ in synthetic substitution -2 1 0 -3 3 0 -2 4 2 1 -2 -1 5 so it’s 5. F(x) = x3 - 3x + 3. Find f(-2) using synthetic substitution 3x – 2 is a factor of 6x3 – 25x2 + 2x + 8 find other factors so 3x – 2 = 0 3x = 2 x = 2/3 2/3 6 -25 2 8 4 -14 -8 6 -21 -12 0 6x2 -21x – 12 3(2x2 – 7x – 4) (2x + 1)(x – 4)

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