1 / 10

Analysis of Polynomial Functions: Finding Rational Zeros and Factors

This practice test provides a comprehensive review of methods to find rational zeros and determine factors of polynomial functions. It covers various examples, including cubic and higher-degree polynomials, with step-by-step synthetic division and analysis of complex zeros. Students are encouraged to identify all possible rational zeros and verify if binomials are factors of given polynomials. Key exercises include exploring the relationships between coefficients and roots, applying the Rational Root Theorem, and simplifying polynomial equations.

wright
Télécharger la présentation

Analysis of Polynomial Functions: Finding Rational Zeros and Factors

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 Page 9 (back) 5) 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 Page 9 (front) 4) List all possible rational zeros of this polynomial:

  4. Page 9 back2) Determine if the binomial is a factor of the polynomial P(x) = x3 + 5x2 +7x + 9; x + 1 -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. Page 9 (back) 4) x3 + x2 – x+ 15 = 0; is 1 – 2i is a zero? 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 -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. Page 9 (back) 9) 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. Page 9 (front) 9) 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 0 (-1 + i )(-2 – 2i) -2 + 2 i - 2 i - 2 i2 -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 – 3 ) ( x – 2 ) ( x + 2) y 5 x -5 -5 5 Page 9 (front) 7) Find all zeros

  9. Page 9 (back) Find all zeros5x4 – 31x3 + 11x2 – 31x + 6 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)

  10. Page 9 (front) 8) Find all zeros 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