1 / 12

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

bary
Download Presentation

Unit 3 Practice Test

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related