1 / 11

Solving Polynomial Equations

Solving Polynomial Equations. PPT 5.3.2. Factor Polynomial Expressions. In the previous lesson, you factored various polynomial expressions. Such as: x 3 – 2x 2 = x 4 – x 3 – 3x 2 + 3x = = =.

krystyn
Download Presentation

Solving Polynomial Equations

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. Solving Polynomial Equations PPT 5.3.2

  2. Factor Polynomial Expressions In the previous lesson, you factored various polynomial expressions. Such as: x3 – 2x2 = x4 – x3 – 3x2 + 3x = = = Grouping – common factor the first two terms and then the last two terms. Refer to 5.2.2 in Lesson 2 to review which strategy is required for each question. Common Factor x2(x – 2) x(x3 – x2 – 3x + 3) x[x2(x – 1) – 3(x – 1)] Common Factor x(x2 – 3)(x – 1)

  3. Solving Polynomial Equations The expressions on the previous slide are now equations: y = x3 – 2x2 and y = x4 – x3 – 3x2 +3x To solve these equations, we will be solving for x when y = 0.

  4. Solve y = x3 – 2x2 0 = x3 – 2x2 0 = x2(x – 2) x2 = 0 or x – 2 = 0 x = 0 x = 2 Therefore, the roots are 0 and 2. Let y = 0 Common factor Separate the factors and set them equal to zero. Solve for x

  5. Solve Let y = 0 y = x4 – x3 – 3x2 + 3x 0 = x4 – x3 – 3x2 + 3x 0 = x(x3 – x2 – 3x + 3) 0 =x[x2(x – 1) – 3(x – 1)] 0 = x(x – 1)(x2 – 3) x = 0 or x – 1 = 0 or x2 – 3 = 0 x = 0 x = 1 x = Therefore, the roots are 0, 1 and ±1.73 Common factor Group Separate the factors and set them equal to zero. Solve for x

  6. What are you solving for? In the last two slides we solved for x when y = 0, which we call the roots. But what are roots? If you have a graphing calculator follow along with the next few slides to discover what the roots of an equation represent.

  7. What are roots? Press the Y= button on your calculator. Type x3 – 2x2

  8. Press the GRAPH button. Look at where the graph is crossing the x-axis. The x-intercepts are 0 and 2. If you recall, when we solved for the roots of the equation y = x3 – 2x2, we found them to be 0 and 2. Don’t forget, we also put 0 in for y, so it makes sense that the roots would be the x-intercepts.

  9. Use your graphing calculator to graph the other equation we solved, y = x4 – x3 – 3x2 + 3x As you would now expect, the roots that we found earlier, 0, 1 and ±1.73, are in fact the x-intercepts of the graph.

  10. The Quadratic Formula For equations in quadratic form: ax2 + bx + c = 0, we can use the quadratic formula to solve for the roots of the equation. This equation is normally used when factoring is not an option.

  11. Remember, the root 0 came from an earlier step. Using the Quadratic Formula Solve the following cubic equation: y = x3 + 5x2 – 9x 0 = x(x2 + 5x – 9) x = 0 x2 + 5x – 9 = 0 We can, however, use the quadratic formula. Can this equation be factored? We still need to solve for x here. Can this equation be factored? YES it can – common factor. No. There are no two integers that will multiply to -9 and add to 5. a = 1 b = 5 c = -9 Therefore, the roots are 0, 6.41 and -1.41.

More Related