1 / 12

Algebra II Section 2.7: Polynomial Zeros and Solutions

Learn how to find the number of solutions and zeros of polynomial equations and functions. Use a calculator or synthetic division to find rational or complex zeros, and factor the polynomial into linear factors.

tmeagan
Download Presentation

Algebra II Section 2.7: Polynomial Zeros and Solutions

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. Algebra IISect 2.7 Please log on to your computers.

  2. Write a formula for the volume of the box below:

  3. Write a formula for the volume of the box below: If the volume is equal to 2730 cubic units, what are the dimensions of the box?

  4. ANSWER 4 GUIDED PRACTICE How many solutions does the equation x4 + 5x2– 36 = 0 have?

  5. ANSWER 4 Use A Calculator! GUIDED PRACTICE How many solutions does the equation x4 + 5x2– 36 = 0 have? What are they?

  6. PRACTICE Find the number of solutions or zeros How many zeros does the function f (x) = x4 – 8x3 + 18x2 – 27have?

  7. Use A Calculator! PRACTICE How many zeros does the function f (x) = x4 – 8x3 + 18x2 – 27have? What are they?

  8. PRACTICE Find the number of solutions or zeros How many zeros does the function f (x) = x4 – 8x3 + 18x2 – 27have? What are the solutions? SOLUTION Because f (x) = x4– 8x3 + 18x2– 27 is a polynomial function of degree 4, it has four zeros. (The zeros are – 1, 3, 3, and 3.)

  9. PRACTICE Find the number of solutions or zeros How many solutions does the equation x3 + 5x2 + 4x + 20 = 0have? SOLUTION Because x3 + 5x2 + 4x + 20 = 0 is a polynomial equation of degree 3,it has three solutions. (The solutions are – 5, – 2i, and 2i.)

  10. STEP 1 Find the rational zeros of f. Because fis a polynomial function of degree 5, it has 5 zeros. The possible rational zeros are + 1, + 2, + 7, and +14. Using synthetic division, you can determine that – 1 is a zero repeated twice and 2 is also a zero. STEP 2 Write f (x) in factored form. Dividing f (x) by its known factors x + 1, x + 1, and x – 2 gives a quotient of x2– 4x + 7. Therefore: EXAMPLE No Calculator! Find all zeros of f (x) = x5 – 4x4 + 4x3 + 10x2 – 13x – 14. SOLUTION f (x) = (x + 1)2(x – 2)(x2 – 4x + 7)

  11. STEP 3 Find the complex zeros of f . Use the quadratic formula to factor the trinomial into linear factors. f(x) = (x + 1)2(x – 2) x – (2 + i 3 ) x – (2 – i 3 ) ANSWER The zeros of f are – 1, – 1, 2, 2 + i 3 , and2 – i 3. EXAMPLE Find the zeros of a polynomial function

  12. Zeros of f are 1, 1, – 2, 1 + i 2, and 1 – i 2 GUIDED PRACTICE Find all zeros of the polynomial function. 1. f (x) = x3 + 7x2 + 15x + 9 ANSWER The zeros of f are – 1, −3, and – 3. 2. f (x) = x5– 2x4 + 8x2– 13x + 6 ANSWER

More Related