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)

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

product of roots = k if ‘n’ is even a = - k of ‘n’ is odd 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

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 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

Example 4: Find a 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

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!