Presentation on theme: "General Results for Polynomial Equations Lesson 2.7"— Presentation transcript:
1General Results for Polynomial Equations Lesson 2.7 The Fundamental Theorem of AlgebraIn the complex number system consisting of all realand imaginary numbers, if P(x) is a polynomial ofdegree ‘n’ (n>0) with complex coefficients, then theequation P(x) = 0 has exactly ‘n’ roots (providing adouble root is counted as ‘2’ roots, a triple root as ‘3’roots, etc)The Complex Conjugates TheoremIf P(x) is a polynomial with real coefficients, and‘a+bi’ is an imaginary root, then automatically ‘a-bi’ mustalso be a root!!!! (and vice-versa)
2Irrational Roots Theorem -- Suppose P(x) is a polynomial with rational coefficientsAnd ‘a’ and ‘b’ are rational numbers, such that √b isIrrational. If ‘a + √b ‘ is a root of the equation P(x) = 0Then ‘a - √b’ is also a root. (and vice versa)Odd Degree Polynomial TheoremIf P(x) is a polynomial of odd degree (1,3,5,7,…) withreal coefficients, then the equation P(x) = 0 has atleast one real root!!!! (For instance P(x) = x3 + …, or P(x) = x5 + …, etc)Theorem 5For the equation axn + bxn-1 + … + k = 0, with k ≠ 0Then:sum of roots = - ba
3product of roots = k if ‘n’ is even a= - k of ‘n’ is oddExample: 2x3 – 5x2 – 3x + 9 = 0 What can you identifyabout this equation.1st: Because this is an odd polynomial -- has at leastone real root.2nd: Sum of the roots = - b = - (-5) = 5aWhich means r1 + r2 + r3 = 52
43rd: Product of roots: (since ‘n’ is odd) = - k = - 9 a 2 Which means r1(r2)(r3) = - 92Example 2: If a 6th degree polynomial equation withrational coefficients has √5 as a root,what is another root?Dah 2 - √5Example 3: Find a quadratic equation with roots: i(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 = ka aSoooo 4 = - b and = kGet a common denominator on both sum and product = - b & 13 = ka aSooo by comparing these two fractions we can reason out that a = 1, b = - 4 & k = 13Therefore our equation is: 1x2 - 4x + 13 = 0
5Example 4: Find a cubic equation with integral coefficients and roots 1 - √6 and - 32So r1 = 1 - √6 , r2 = 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 = - 5now let 2 = - b and let – 5 = k ; so we can summarize a = 1, b = -2 & k = - 5a aSo the quadratic factor that would give us the two conjugate roots would be:(1x2 -2x – 5) Now take r2 = 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 integralcoefficients that has no ‘quadratic’ termand i√2 as one of its roots.Since this is cubic, there has to be ‘3’ roots, and because of an earliertheorem recently given we should all know that r2 = 3 - i√2Right 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
60x2 soo b = 0.Hmmm that means sum of my roots would be - b but since b = 0 , then the sum = 0!aSooo r1 + r2 + r3 = 0 Which means (3 + i√2) + (3 - i√2) + r3 = 0 r3 = 0 r3 = - 6 x = - 6Now 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= = 11so we set 6 = - b and set 11 = k1 a a ( k here because we are just gettinga the quadratic factor here. )Soo a = 1, b = - 6, and k = 11Therefore the quadratic ‘factor’ is (x2 – 6x + 11)Now take the third root and get the factor (x + 6) multiply these two factorstogether (x2 – 6x + 11)(x + 6) -> be careful and distribute and get: ????????? You do the work!