Download presentation

Published byJoseph Ramirez Modified over 3 years ago

1
**Aim: How do we solve polynomial equations using factoring?**

Do Now: Write the expression (x + 1)(x + 2)(x + 3) as a polynomial in standard form. (x + 1)(x + 2)(x + 3) FOIL first two factors (x2 + 3x + 2)(x + 3) Multiply by distribution of resulting factors x3 + 3x2 + 3x2 + 9x + 2x + 6 Combine like terms x3 + 6x2 + 11x + 6 cubic expression is standard form

2
**Solving Polynomials by Factoring**

Factor: 2x3 + 10x2 + 12x GCF 2x(x2 + 5x + 6x) Factor trinomial 2x(x + 3)(x + 2) Solve: 2x3 + 10x2 + 12x = 0 2x(x2 + 5x + 6x) = 0 2x(x + 3)(x + 2) = 0 2x = 0 (x + 3) = 0 (x + 2) = 0 Set factors equal to zero (Zero Product Property) x = 0, -2, -3

3
**Cubic equation in Standard form**

Graphical Solutions Solve: 2x3 + 10x2 + 12x = 0 2x(x + 3)(x + 2) = 0 x = 0, -2, -3 Cubic equation in Standard form ax3 + bx2 + cx + d = y y-intercept

4
**Find the zeros of y = (x – 2)(x + 1)(x + 3) **

Model Problem Find the zeros of y = (x – 2)(x + 1)(x + 3) Zeros/roots/x-intercepts are found at y = 0 (x-axis) 0 = (x – 2)(x + 1)(x + 3) (x – 2) = 0 (x + 1) = 0 (x + 3) = 0 Set factors equal to zero x = -3, -1, 2 y-intercept? (-2) (1) (3) = -6 is the product of last terms of binomial factors y = x3 + 2x2 – 5x – 6

5
**Synthetic Division & Factors**

Divide x3 – x by x + 1 using synthetic division -1 Since there is no remainder x + 1 is a factor of x3 – x2 + 2 -1 2 -2 1 -2 2 x2 – 2x + 2 quotient remainder Remainder Theorem (x2 – 2x + 2)(x + 1) = x3 – x2 + 2 Solve: x3 – x2 + 2 = 0 (x2 – 2x + 2)(x + 1) = 0 (x2 – 2x + 2) = 0 (x + 1) = 0 Set factors equal to zero Quadratic Formula x = -1 x = 1 ± i

6
**Synthetic Division, Factors and Graphing**

Solve: x3 – x2 + 2 = 0 (x2 – 2x + 2)(x + 1) = 0 (x2 – 2x + 2) = 0 (x + 1) = 0 Set factors equal to zero Quadratic Formula x = -1 x = 1 ± i

7
**Model Problem solve 2x3 + 11x2 + 18x + 9 = 0**

Use synthetic division to show that x + 3 is a factor of y = 2x3 + 11x2 + 18x + 9 then solve 2x3 + 11x2 + 18x + 9 = 0 -3 -6 -15 -9 2 5 3 2x3 + 11x2 + 18x + 9 = 0 2x2 + 5x + 3 (2x2 + 5x + 3)(x + 3) = 0 (2x + 3)(x + 1)(x + 3) = 0 (2x + 3) = 0 (x + 1) = 0 (x + 3) = 0 x = (-3, -3/2, -1) 7

8
**Factor by Grouping x3 – 2x2 – 3x + 6 (x3 – 2x2) – (3x – 6)**

Group terms (x3 – 2x2) – (3x – 6) Factor Groups x2(x – 2) – 3(x – 2) Distributive Property (x2 - 3)(x – 2)

9
**Model Problem Factor: 4x3 – 6x2 + 10x – 15 (4x3 – 6x2) + (10x – 15)**

Group terms (4x3 – 6x2) + (10x – 15) Factor Groups 2x2(2x – 3) + 5(2x – 3) Distributive Property (2x2 + 5)(2x – 3) Factor: x3 – 2x2 – 4x + 8 (x3 – 2x2) – (4x – 8) x2(x – 2) + 4(x – 2) (x2 + 4)(x – 2) 9

10
**Model Problem Solve: x3 – 3x2 – 3x + 9 = 0 (x3 – 3x2) – (3x – 9) = 0**

Group x2(x – 3) – 3(x – 3) = 0 Factor (x2 – 3)(x – 3) = 0 (x – 3) = 0 x = 3 (x2 – 3) = 0 x =

11
**Solving Quadratic Form Equations**

x4 – 3x2 + 2 = 0 u = x2 u2 – 3u + 2 = 0 u2 – 3u + 2 = 0 Quadratic form (u – 1)(u – 2) = 0 Factor (x2 – 1)(x2 – 2) = 0 Substitute (x – 1)(x + 1)(x2 – 2) = 0 Factor (x – 1) = 0 x = 1 Set factors = 0 & solve for x (x + 1) = 0 x = -1 (x2 + 2) = 0 x =

12
Model Problem Solve: x4 – x2 = 12

Similar presentations

Presentation is loading. Please wait....

OK

Copyright © Cengage Learning. All rights reserved.

Copyright © Cengage Learning. All rights reserved.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on structural changes in chromosomes Ppt on paintings and photographs related to colonial period Ppt on relations and functions for class 11th economics Ppt on state of indian economy Ppt on life history of bill gates Ppt on ip address classes Ppt on 2nd world war video Ppt on social contract theory of government Ppt on solar system and earth Ppt on introduction to object-oriented programming php