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Unit 3 Practice Test

Unit 3 Practice Test. Review. 5x 4 – 31x 3 + 11x 2 – 31x + 6 p  1, 2, 3, 6 q  1, 5 p  1, 2, 3, 6, 1/5, 2/5, 3/5, 6/5 q . 1a) List all possible rational zeros of this polynomial: . -2x 3 + 5x 2 + 6 x + 18 p  1, 2, 3, 6, 9, 18 q  1, 2

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Unit 3 Practice Test

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  1. Unit 3 Practice Test Review

  2. 5x4– 31x3+ 11x2 – 31x + 6 p  1, 2, 3, 6 q  1, 5 p  1, 2, 3, 6, 1/5, 2/5, 3/5, 6/5 q 1a) List all possible rational zeros of this polynomial:

  3. -2x3 + 5x2 + 6x + 18 p  1, 2, 3, 6, 9, 18 q  1, 2 p  1, 2, 3, 6, 9, 18, 1/2, 3/2,9/2 q 1b) List all possible rational zeros of this polynomial:

  4. 2) Use Synthetic Division to determine if x + 1is a factor of x3 + 5x2 +7x + 9 SHOW WORK NEATLY AND EXPLAIN YOUR ANSWER -1 1 5 7 9 -1 - 4 - 3 1 4 3 6 No when synthetically dividing with -1, the remainder is 6 not 0; so x + 1 is not a factor.

  5. 3)Determine if 1 – 2i is a zero x3+ x2– x+ 15SHOW WORK NEATLY AND EXPLAIN YOUR ANSWER 1 – 2i| 1 1 -1 15 1 – 2i 1 2 – 2i (1 – 2i )(2 – 2i ) 2 – 2i – 4i + 4i 2 -2 – 6i 1 – 2i| 1 1 -1 15 1 – 2i -2 – 6i 1 2 – 2i -3 – 6i (1 – 2i )(-3 – 6i) -3 – 6i + 6i + 12i2 -3 – 12 = -15 1 – 2i| 1 1 -1 15 1 – 2i -2 – 6i -15 1 2 – 2i -3 – 6i 0 Yes, when synthetically dividing with 1 – 2i, the remainder is zero; so 1 – 2i is a zero.

  6. 4) If -1/3 is a zero of h(x) = 3x3 – 2x2 – 61x – 20 , find the other zeros. 3x3 – 2x2 – 61x – 20 -1/3 | 3 -2 -61 -20 -1 1 20 3 -3 -60 0 3x2 – 3x – 60 = 0 3(x2 – x – 20) = 0 3(x – 5)(x + 4) x = 5, -4

  7. 5) x4 + 2x3 – 4x–4; -1 + i is a zero -1 + i| 1 2 0 -4 -4 -1 + i 1 1 +i (-1 + i )(1 + i ) -1 – i + i + i2 -2 -1 + i| 1 2 0 -4 -4 -1 + i -2 2 – 2i 4 1 1 +I -2 -2 – 2i (-1 + i )(-2 – 2i) 2 + 2 i - 2 i - 2 i2 2 + 2 = 4 -1 - i| 1 1 + i -2 -2 – 2i 0 -1 – i 0 2 + i 1 0 -2 0 x 2 – 2 = 0 x = ±2

  8. x4– x3– 10x2 + 4x + 24 -2 | 1 -1 -10 4 24 ↓ -2 6 8 -24 1 -3 -4 12 0 x3 – 3 x2 – 4x + 12 = 0 x2 ( x – 3 ) – 4 (x – 3 ) ( x – 3 ) (x2 – 4 ) (x + 2 )(x – 3 ) ( x – 2 ) ( x + 2) y 5 x -5 -5 5 6A) Graph (x + 2 )(x – 3 ) ( x – 2 ) ( x + 2)

  9. x4– 20x2+ 64 (x2 – 4)(x2 – 16) (x – 2)(x + 2)(x – 4)(x + 4) x = 2, -2, 4, -4 y 5 x -5 -5 5 6B) Graph

  10. 1)Find all zeros and factor5x4 – 31x3 + 11x2 – 31x + 6 From 1a p  1, 2, 3, 6, 1/5, 2/5, 3/5, 6/5 q 6  5 -31 11 -31 6 30 -6 30 -6 1/5 5 -1 5 -1 0 1 0 1 5 0 5 0 5x2 + 5 = 0 5(x2 + 1) = 0 x2 = -1 x = ± i x = 6, 1/5, ± i5(x + i)(x – i)(x – 6)(x – 1/5)

  11. 1)Find all zeros and factor5x4 – 31x3 + 11x2 – 31x + 6 6  5 -31 11 -31 6 30 -6 30 -6 5 -1 5 -1 0 5x3 – 1x2+ 5x – 1 x2(5x – 1) + 1(5x – 1) (5x – 1)(x2 + 1) x2 + 1 = 0 x2= -1 x = ± i x = 6, 1/5, ± i 5(x + i)(x – i)(x – 6)(x – 1/5) 6  5 -31 11 -31 6 30 -6 30 -6 1/5 5 -1 5 -1 0 1 0 1 5 0 5 0 5x2 + 5 = 0 5(x2 + 1) = 0 x2 = -1 x = ± i x = 6, 1/5, ± i 5(x + i)(x – i)(x – 6)(x – 1/5)

  12. 2) Find all zeros and factor 18x3 + 3x2 – 7x – 2 p  1, 2 q  1, 2 , 3, 6, 9, 18 p  1, 2, ½, 1/3, 2/3, 1/6, q 1/9, 2/9, 1/18 18x3 + 3x2 – 7x – 2 -½ | 18 3 -7 -2 ↓ -9 3 2 18 -6 -4 0 18x2 – 6x – 4 = 0 2(9x2 – 3x – 2) = 0 2(3x + 1)(3x – 2) x = -1/2, -1/3, 2/3 2(3x + 1)(3x – 2)(x + ½)

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