Let’s Review Synthetic Division

1. Let’s Review Synthetic Division

2. Ex. 1 (4x3 + x + 7)  (x – 2) Set = to zero & solve for x 2

3. Ex. 1 (4x3 + x + 7)  (x – 2) add down multiply by the box number 2 41 Remainder Quotient

4. Ex. 2 You try… (x3 + 6x2 – x – 30)  (x – 2) 30

5. Ex. 3 (2x3 + 7x2 – 5)  (x + 3)

6. Higher Degree Polynomial Functions 2.6

7. An example of a polynomial function… f(x) = 6x4 + x3 – 21x2 – 15x + 36 Leading coefficient Degree (highest exponent) (in front)

8. An example of a polynomial function… f(x) = 6x4 + x3 – 21x2 – 15x + 36 When we solve a polynomial function, we are looking for the numbers that make the equation equal to zero. “root” “zero” and “solution” all mean the same thing

9. From Last week… f(x) = x2 + x – 6 To solve this equation… x2 + x – 6 = 0 Set equal to zero (x + 3)(x – 2) = 0 Factor Set each factor equal to zero x + 3 = 0 x – 2 = 0 x = - 3 x = 2 solve

10. f(x) = 6x4 + x3 – 21x2 – 15x + 36 • But with larger polynomials, solving by factoring is impractical. • Later we’ll learn new methods for solving a polynomial.

11. How many roots? How many factors? f(x) = x3 + 2x2 – 3x + 12 f(x) = x3 + 2x2 – 3x + 12 3 3 f(x) = x5 + 14x4 – 4 f(x) = x5 + 14x4 – 4 5 5 The degree tells you the number of roots or factors the polynomial will have.

12. Our goal is to be able to find all of the roots of a polynomial.

13. Finding Roots

14. Fundamental Theorem of Algebra -Every polynomial will have at least one linear factor

15. Repeated roots… Ex. 1 P(x) = x2 – 10x + 25 P(x) = (x - 5)(x - 5) x = 5 x = 5 5 has a multiplicity of two If a root occurs k times, it has a multiplicity of k.

16. Repeated roots… f(x) = x2 f(x) = (x+3)(x+3)(x+3) x = 0, 0 x = -3, -3, -3 0 has a multiplicity of two -3 has a multiplicity of three

17. Conjugate pairs -If is a root, then is also a root. Imaginary roots always come in pairs. -If a + biis a root, then a – bi is also a root. Irrational roots always come in pairs.

18. Sometimes you can find all of the roots of a polynomial without doing much work.

19. Suppose a polynomial of degree 3 has 3 – 4i and 9 as roots. Find all of the roots. 3 Ex. 2 Roots Because the degree is 3… 9 we should have 3 roots. 3 – 4i Because one root is 3 – 4i… 3 + 4i then 3 + 4i is also a root. Imaginary roots always come in pairs.

20. Ex. 3 You try… Suppose a polynomial of degree 6 has -2 + 5i, -i, and as roots. Find all of the roots. 6 Because the degree is 6… Roots we should have 6 roots. -2 + 5i -2 – 5i -i Imaginary roots always come in pairs. i Irrational roots always come in pairs.

21. Ex. 4 Find all of the roots. f(x) = x2 – 2x - 8 If quadratic: (x – 4)(x + 2) factor OR quadratic formula

22. Ex. 5 Given a zero, find all the zeros P(x) = x3 – 6x2 + 13x – 20;4 We should get zero for a remainder… 20 4 -8 why? 1 -2 5 Now we have a quadratic, what can we do? Factor OR quadratic formula

23. Ex. 5 Given a zero, find all the zeros P(x) = x3 – 6x2 + 13x – 20;4 , 12i 20 4 -8 1 -2 5

24. Ex. 6 Given a zero, find the other zeros. , -2i P(x) = x4 – 16;2i If 2i is a zero, then…… -2i is a zero. 1 0 0 0 -16 2i -8i 16 -4 2i -4 -8i 0 1 now do synthetic again

25. Ex. 6 Given a zero, find the other zeros. , -2i P(x) = x4 – 16;2i , 2, -2 1 0 0 0 -16 2i -8i 16 -4 2i -4 -8i 0 x2 – 4 = 0 1 0 8i -2i (x – 2)(x + 2)= 0 0 1 0 -4 x = 2 x = -2

26. Notes • Let P be a polynomial function in standard form with integer coefficients. If is a root of P(x), then: • p is a factor of the constant term of P • q is a factor of the leading coefficient of P The Rational Roots Theorem

27. Discussion List all possible rational roots of P

28. Discussion Find all roots of P:

29. Discussion Find all roots of P:

30. Classwork: Page 101 Homework: Page 1-11 ALL