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Download 4.6: The Fundamental Theorem of Algebra
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The Fundamental Theorem of Algebra Section 4.6 beginning on page 198 The Fundamental Theorem of Algebra Essentially, this theorem tells us that the degree of the function is equal to the number of solutions (or zeros) the function has, provided we count repeated real and imaginary solutions. Example 1: How many solutions does each equation have ? a) π₯ 3 + 3π₯ 2 + 16π₯ + 48 = 0 a) 3 b) 4 b) π π₯ = π₯ 4 + 6π₯ 3 + 12π₯ 2 + 8π₯ Finding the Zeros of a Polynomial Example 2: Find all zeros of π π₯ = π₯ 5 + π₯ 3 β 2π₯ 2 β 12π₯ β 8 Steps: β’ Use the rational root theorem to find possible zeros. β’ Use synthetic division to test them (The fundamental theorem of algebra tells us how many we should have) β’ Write the polynomial in factored form. (We will test together on the board) β’ Find the zeros from the factorization. ** Once you find one, divide into the NEW quotient. Once you get down to a quadratic ±1, ±2, ±4, ±8 you can factor the rest if possible) We determined that -1 is a factor twice and 2 is a factor and the remaining quotient is π₯ 2 + 4. π π₯ = π₯ + 1 2 (π₯ β 2)(π₯ 2 + 4) π₯2 + 4 = 0 π₯ 2 = β4 π₯ = β4 π₯ = β1, π₯ = 2, π₯ = ±2π The Complex Conjugates Theorem ** This is nearly identical to the irrational conjugates theorem from section 4.5** If π + ππ is an imaginary zero of a polynomial π, then π β ππ is also an imaginary zero of π. Using Zeros to Write a Polynomial Example 3: Write a polynomial f of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 2 and 3 + π. Because of the complex conjugates theorem we know that 3 β π is also a zero. π π₯ = π₯ β 2 π₯ β 3 + π [π₯ β 3 β π ] π π₯ = π₯ β 2 (π₯ β 3) β π [(π₯ β 3) + π] π π₯ = π₯ β 2 [ π₯ β 3 2 β π 2] π π₯ = π₯ β 2 [(π₯ 2 β 6π₯ + 9) + 1] π π₯ = π₯ β 2 (π₯ 2 β 6π₯ + 10) π π₯ = π₯ 3 β 6π₯ 2 + 10π₯ β 2π₯ 2 + 12π₯ β 20 π π₯ = π₯ 3 β 8π₯ 2 + 22π₯ β 20 Step 1: Write the polynomial in factored form. Step 2: Regroup the terms to make the second two factors a special sum and difference. Step 3: Multiply out and simplify that special product. (Remember π 2 = β1) Step 3: Multiply the rest out and simplify . Descartesβs Rule of Signs The number of positive real zeros of a function π is equal to the number of changes in sign of the coefficients of π(π₯) or is less than this by an even number. The number of negative real zeros of a function π is equal to the number of changes in sign of the coefficients of π(βπ₯) or is less than this by an even number. With this information and the fundamental theorem of algebra (which tells us the total number of zeros), we can identify all of the possible combinations of positive real zeros, negative real zeros, and imaginary zeros. Using Descartesβs Rule of Signs Example 4: Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for π π₯ = π₯ 6 β 2π₯ 5 + 3π₯ 4 β 10π₯ 3 β 6π₯ 2 β 8π₯ β 8. π π₯ = π₯ 6 β 2π₯ 5 + 3π₯ 4 β 10π₯ 3 β 6π₯ 2 β 8π₯ β 8 π βπ₯ = (βπ₯)6 β2 βπ₯ 5 + 3 βπ₯ 4 β 10 βπ₯ 3 β 6 βπ₯ π βπ₯ = π₯ 6 + 2π₯ 5 + 3π₯ 4 + 10π₯ 3 β 6π₯ 2 + 8π₯ β 8 Positive Real 3 Negative Real Imaginary Total 6 3 0 3 1 2 6 1 3 2 6 1 1 4 6 3 sign changes ο 3 or 1 positive real zeros. 2 β 8(βπ₯) β 8 3 sign changes ο 3 or 1 negative real zeros. We know that there are 6 solutions (the degree of the polynomial is 6) Monitoring Progress 1) How many solutions does the equation π₯ 4 + 7π₯ 2 β 144 = 0 have? 2) How many zeros does the function π π₯ = π₯ 3 β 5π₯ 2 β 8π₯ + 48 have? 4 3 Find all zeros of the polynomial function. β3, β2, and β 2 3) π π₯ = π₯ 3 + 7π₯ 2 + 16π₯ + 12 4) π π₯ = π₯ 5 β 3π₯ 4 + 5π₯ 3 β π₯ 2 β 6π₯ + 4 β1,1,1,1 + π 3, and 1 β π 3 Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1 and , and the given zeros. 5) -1, 4π 6) 3, 1 + π 5 7) 2, 1 β 3π 8) 2, 2π, 4 β 6 5) π₯ 3 + π₯ 2 + 16π₯ + 16 7) π₯ 4 β 2π₯ 3 + 8π₯ 2 + 4π₯ β 20 6) π₯ 3 β 5π₯ 2 + 12π₯ β 18 8) π₯ 5 β 10π₯ 4 β 30π₯ 3 β 60π₯ 2 + 104π₯ β 80 Monitoring Progress Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function. 1) π π₯ = π₯ 3 + 9π₯ β 25 1) 2) 2) π π₯ = 3π₯ 4 β 7π₯ 3 + π₯ 2 β 13π₯ + 8 Positive Real Negative Real Imaginary Total 1 0 3 Positive Real Negative Real Imaginary Total 4 0 0 4 2 0 2 4 0 0 4 4 2