General Results for Polynomial Equations Lesson 2.7 The Fundamental Theorem of Algebra

1. General Results for Polynomial Equations Lesson 2.7 The Fundamental Theorem of Algebra In the complex number system consisting of all real and imaginary numbers, if P(x) is a polynomial of degree ‘n’ (n>0) with complex coefficients, then the equation P(x) = 0 has exactly ‘n’ roots (providing a double root is counted as ‘2’ roots, a triple root as ‘3’ roots, etc) The Complex Conjugates Theorem If P(x) is a polynomial with real coefficients, and ‘a+bi’ is an imaginary root, then automatically ‘a-bi’ must also be a root!!!! (and vice-versa)

2. Irrational Roots Theorem -- Suppose P(x) is a polynomial with rational coefficients And ‘a’ and ‘b’ are rational numbers, such that √b is Irrational. If ‘a + √b ‘ is a root of the equation P(x) = 0 Then ‘a - √b’ is also a root. (and vice versa) Odd Degree Polynomial Theorem If P(x) is a polynomial of odd degree (1,3,5,7,…) with real coefficients, then the equation P(x) = 0 has at least one real root!!!! (For instance P(x) = x3 + …, or P(x) = x5 + …, etc) Theorem 5 For the equation axn + bxn-1 + … + k = 0, with k ≠ 0 Then: sum of roots = - b a

3. product of roots = k if ‘n’ is even a = - k of ‘n’ is odd a Example: 2x3 – 5x2 – 3x + 9 = 0 What can you identify about this equation. 1st: Because this is an odd polynomial -- has at least one real root. 2nd: Sum of the roots = - b = - (-5) = 5 a 2 2 Which means r1 + r2 + r3 = 5 2

4. 3rd: Product of roots: (since ‘n’ is odd) = - k = - 9 a 2 Which means r1(r2)(r3) = - 9 2 Example 2: If a 6th degree polynomial equation with rational coefficients has 2 + √5 as a root, what is another root? Dah  2 - √5 Example 3: Find a quadratic equation with roots: 2 + 3i (Hint: Find the sum of these ‘two’ roots:  (2 + 3i) + (2 – 3i) = ?? (Now: Find the ‘product of these ‘two’ roots  (2 + 3i)(2 – 3i) = ??? set: sum = - b and set: product = k a a Soooo 4 = - b and 13 = k a a Get a common denominator on both sum and product  4 = - b & 13 = k 1 a 1 a Sooo  by comparing these two fractions we can reason out that a = 1, b = - 4 & k = 13 Therefore our equation is: 1x2 - 4x + 13 = 0

5. Example 4: Finda cubic equation with integral coefficients and roots 1 - √6 and - 3 2 So r1 = 1 - √6 , r2 = - 3 and r3 = ?? (1 + √6) Take the two conjugate roots and find their sum and product: (1 - √6) + (1 + √6 ) = 2 ; (1 - √6 )(1 + √6 ) = 1 – 6 = - 5 now let 2 = - b and let – 5 = k ; so we can summarize a = 1, b = -2 & k = - 5 1 a 1 a So the quadratic factor that would give us the two conjugate roots would be: (1x2 -2x – 5) Now take r2 = - 3 and let x = - 3  (x+3) Now multiply (1x2 -2x – 5) (x + 3)  take your time and distribute out correctly to get the cubic polynomial! Example 5: Find a cubic equation with integral coefficients that has no ‘quadratic’ term and 3 + i√2 as one of its roots. Since this is cubic, there has to be ‘3’ roots, and because of an earlier theorem recently given we should all know that r2 = 3 - i√2 Right now I am confused as to how to figure out r3. Butt – if this equation has ‘no’ quadratic term that means it would actually be

6. 0x2 soo b = 0. Hmmm that means sum of my roots would be - b but since b = 0 , then the sum = 0! a Sooo  r1 + r2 + r3 = 0 Which means (3 + i√2) + (3 - i√2) + r3 = 0  6 + r3 = 0  -6 -6  r3 = - 6  x = - 6 Now take the two ‘conjugate’ roots we have and find their sum and product. sum = (3 + i√2) + (3 - i√2) = 6 and product = (3 + i√2) (3 - i√2) = 9 – 2i2 = 9 + 2 = 11 so we set 6 = - b and set 11 = k 1 a 1 a ( k here because we are just getting a the quadratic factor here. ) Soo  a = 1, b = - 6, and k = 11 Therefore the quadratic ‘factor’ is  (x2 – 6x + 11) Now take the third root and get the factor (x + 6)  multiply these two factors together  (x2 – 6x + 11)(x + 6) -> be careful and distribute and get: ????????? You do the work!