1. Copyright © Cengage Learning. All rights reserved. 0 Precalculus Review

2. Copyright © Cengage Learning. All rights reserved. 0.5 Solving Polynomial Equations

3. 3 Polynomial Equation A polynomial equation in one unknown is an equation that can be written in the form ax n + bx n – 1 + ··· + rx + s = 0 where a, b,..., r, and s are constants. We call the largest exponent of x appearing in a nonzero term of a polynomial the degree of that polynomial.

4. 4 Solving Polynomial Equations Quick Examples 1.3x + 1 = 0 has degree 1 because the largest power of x that occurs is x = x 1. Degree 1 equations are called linear equations. 2.x 2 – x – 1 = 0 has degree 2 because the largest power of x that occurs is x 2. Degree 2 equations are also called quadratic equations, or just quadratics. 3.x 3 = 2x 2 + 1 is a degree 3 polynomial (or cubic) in disguise. It can be rewritten as x 3 – 2x 2 – 1 = 0, which is in the standard form for a degree 3 equation. 4.x 4 – x = 0 has degree 4. It is called a quartic.

5. 5 Solution of Linear Equations

6. 6 By definition, a linear equation can be written in the form ax + b = 0. a and b are fixed numbers with a  0.

7. 7 Solution of Quadratic Equations

8. 8 By definition, a quadratic equation has the form ax 2 + bx + c = 0. The solutions of this equation are also called the roots of ax 2 + bx + c. We’re assuming that you saw quadratic equations somewhere in high school but may be a little hazy about the details of their solution. There are two ways of solving these equations—one works sometimes, and the other works every time. a, b, and c are fixed numbers with a  0.

9. 9 Solution of Quadratic Equations Solving Quadratic Equations by Factoring (works sometimes) If we can factor a quadratic equation ax 2 + bx + c = 0, we can solve the equation by setting each factor equal to zero. Quick Example x 2 + 7x + 10 = 0 (x + 5)(x + 2) = 0 x + 5 = 0 or x + 2 = 0 Solutions: x = –5 and x = –2 Factor the left-hand side. If a product is zero, one or both factors is zero.

10. 10 Solution of Quadratic Equations Test for Factoring The quadratic ax 2 + bx + c, with a, b, and c being integers (whole numbers), factors into an expression of the form (rx + s)(tx + u) with r, s, t, and u integers precisely when the quantity b 2 – 4ac is a perfect square. (That is, it is the square of an integer.) If this happens, we say that the quadratic factors over the integers.

11. 11 Solution of Quadratic Equations Quick Example x 2 + x + 1 has a = 1, b = 1, and c = 1, so b 2 – 4ac = –3, which is not a perfect square. Therefore, this quadratic does not factor over the integers. Solving Quadratic Equations with the Quadratic Formula (works every time) The solutions of the general quadratic ax 2 + bx + c = 0 (a ≠ 0) are given by

12. 12 Solution of Quadratic Equations We call the quantity  = b 2 – 4ac the discriminant of the quadratic (  is the Greek letter delta), and we have the following general rules: If  is positive, there are two distinct real solutions. If  is zero, there is only one real solution: (Why?) If  is negative, there are no real solutions.

13. 13 Solution of Quadratic Equations Quick Example

14. 14 Solution of Cubic Equations

15. 15 Solution of Cubic Equations By definition, a cubic equation can be written in the form ax 3 + bx 2 + cx + d = 0. a, b, c, and d are fixed numbers and a  0.

16. 16 Solution of Cubic Equations Solving Cubics by Finding One Factor Start with a given cubic equation ax 3 + bx 2 + cx + d = 0. Step 1 By trial and error, find one solution x = s. If a, b, c, and d are integers, the only possible rational solutions are those of the form s = ±(factor of d)/(factor of a). Step 2 It will now be possible to factor the cubic as ax 3 + bx 2 + cx + d = (x – s)(ax 2 + ex + f ) = 0 To find ax 2 + ex + f, divide the cubic by x – s, using long division.

17. 17 Solution of Cubic Equations Step 3 The factored equation says that either x – s = 0 or ax 2 + ex + f = 0. We already know that s is a solution, and now we see that the other solutions are the roots of the quadratic. Note that this quadratic may or may not have any real solutions, as usual.

18. 18 Solution of Cubic Equations Quick Example To solve the cubic x 3 – x 2 + x – 1 = 0, we first find a single solution. Here, a = 1 and d = –1. Because the only factors of ±1 are ±1, the only possible rational solutions are x = ±1. By substitution, we see that x = 1 is a solution. Thus, (x – 1) is a factor. Dividing by (x – 1) yields the quotient (x 2 + 1). Thus, x 3 – x 2 + x – 1 = (x – 1)(x 2 + 1) = 0 so that either x – 1 = 0 or x 2 + 1 = 0.

19. 19 Solution of Cubic Equations Because the discriminant of the quadratic x 2 + 1 is negative, we don’t get any real solutions from x 2 + 1 = 0, so the only real solution is x = 1. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. One real solution (as in the Quick Example above) 2. Two real solutions (try, for example, x 3 + x 2 – x – 1 = 0) 3. Three real solutions (see the next example)

20. 20 Example 1 – Solving a Cubic Solve the cubic 2x 3 – 3x 2 – 17x + 30 = 0. Solution: First we look for a single solution. Here, a = 2 and d = 30. The factors of a are  1 and  2, and the factors of d are  1,  2,  3,  5,  6,  10,  15, and  30. This gives us a large number of possible ratios:  1,  2,  3,  5,  6,  10,  15,  30,  1/2,  3/2,  5/2,  15/2. Undaunted, we first try x = 1 and x = –1, getting nowhere.

21. 21 Example 1 – Solution So we move on to x = 2, and we hit the jackpot, because substituting x = 2 gives 16 – 12 – 34 + 30 = 0. Thus, (x – 2) is a factor. Dividing yields the quotient 2x 2 + x – 15. Here is the calculation: cont’d

22. 22 Example 1 – Solution Thus, 2x 3 – 3x 2 – 17x + 30 = (x – 2)(2x 2 + x – 15) = 0. Setting the factors equal to zero gives either x – 2 = 0 or 2x 2 + x – 15 = 0. We could solve the quadratic using the quadratic formula, but, luckily, we notice that it factors as 2x 2 + x – 15 = (x + 3)(2x – 5). Thus, the solutions are x = 2, x = –3 and x = 5/2. cont’d

23. 23 Solution of Higher-Order Polynomial Equations

24. 24 Solution of Higher-Order Polynomial Equations Logically speaking, our next step should be a discussion of quartics, then quintics (fifth degree equations), and so on forever. Well, we’ve got to stop somewhere, and cubics may be as good a place as any. On the other hand, since we’ve gotten so far, we ought to at least tell you what is known about higher order polynomials.

25. 25 Solution of Higher-Order Polynomial Equations Quartics Just as in the case of cubics, there is a formula to find the solutions of quartics. Quintics and Beyond All good things must come to an end, we’re afraid. It turns out that there is no “quintic formula.” In other words, there is no single algebraic formula or collection of algebraic formulas that gives the solutions to all quintics.