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5.7 Apply the Fundamental Theorem of Algebra Goal Classify the zeros of polynomial functions. Your Notes VOCABULARY Repeated Solution For the equation f(x) = 0, k is a repeated solution if and only if the factor (x k) has a degree greater than 1 when f is factored completely. THE FUNDAMENTAL THEOREM OF ALGEBRA Theorem: If f(x) is a polynomial of degree n where n __ 0, then the equation f(x) = 0 has at least __one__ solution in the set of complex numbers. Corollary: If f(x) is a polynomial of degree n, then the equation f(x) = 0 has exactly _n_ solutions provided each solution repeated twice is counted as _2_ solutions, each solution repeated three times is counted as _3_ solutions and so on. Example 1 Find the number of solutions or zeros Find the number of solutions or zeros for each equation or function. a. Because x3 3x2 + 9x 27 = 0 is a __third__ degree polynomial equation, it has __three__ solutions. b. Because f(x) = x4 + 6x3 32x is a __fourth__ degree polynomial function, it has __four__ zeros. Checkpoint Complete the following exercise. 1. State the number of zeros of f(x) = x3 2x2 9x + 18. three zeros Your Notes Example 2 Find the zeros of a polynomial function Find all zeros of f(x) = x5 5x4 9x3 5x2 8x + 12. Solution 1. Find the rational zeros of f. Because f is a fifth-degree function, it has __five__ zeros. The possible rational zeros are __±1, ±2, ±3, ±4, ±6, and ±12__ . Using synthetic division, you can determine that _2_ is a zero repeated twice and _1_ is also a zero. 2. Write f(x) in factored form. Dividing f by its known factors gives a quotient of __x2 2x + 3__ . So, f(x) = __(x 2)2 (x + 1) (x2 2x + 3)__ 3. Find the complex zeros of f. Use the quadratic formula to factor the trinomial into linear factors. f(x) = ___(x 2)2 (x + 1)[x (1 + i 2 )][x (1 i 2 )]___ The zeros of f are __1, 2, 2, 1 + i 2 , and 1 i 2 __ Checkpoint Find all zeros of the polynomial function. 2. f(x) = x4 7x3 + 13x2 + x 20 1, 4, 2 + i, 2 i COMPLEX CONJUGATES THEOREM If f is a polynomial function with __real__ coefficients, and __a + bi__ is an imaginary zero of f, then __a bi__ also a zero of f. IRRATIONAL CONJUGATES THEOREM Suppose f is a polynomial function with __rational__ coefficients, and a and b are b rational numbers such that b is irrational. If __a + b __ is a zero of f, then __a __ is also a zero of f. Your Notes Example 3 Use zeros to write a polynomial function Write a polynomial function f of least degree that has real coefficients, a leading coefficient of 1, and 2 and 3 + i as zeros. Because the coefficients are real and 3 + i is a zero, __3 i__ must also be a zero. Use the three zeros and the factor theorem to write f(x) as a product of three factors. f(x) = ( _x + 2_ )[x ( _3 + i_ )][x ( _3 i_ )] Factored form = ( _x + 2_ )[ _(x- 3) i_ ][ _(x 3) + i_ ] Regroup terms. = (x+ 2)[(x 3)2 i2] Multiply. = = (x+ 2)[x2 6x+ 9 (1)] (x+ 2)(x2 6x+ 10) Expand, use i2 = 1. Simplify. = x3 6x2 + 10x+ 2x2 12x + 20 Multiply. = x3 4x2 2x + 20 Combine like terms. You can check this result by evaluating f(x) at each of its three zeros. Checkpoint Complete the following exercise. 3. Write a polynomial of least degree that has rational coefficients, a leading coefficient of 1, and 4 and 1 + 6 as zeros. f(x) = x3 6x2 + 3x + 20 DESCARTES' RULE OF SIGNS Let f(x) = anxn + an 1xn 1 + …….+ a2x2 + a1x + a0 be a polynomial function with real coefficients. The number of __positive__ real zeros of f is equal to the number of changes in sign of the coefficients of _f(x)_ or is less than this by an _even_ number. The number of __negative__ real zeros of f is equal to the number of changes in sign of the coefficients of __f(x)__ or is less than this by an __even__ number. Your Notes Example 4 Use Descartes'rule of signs Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for f(x) = 2x5 7x4 + 12x3 + 2x2 + 4x + 6. Solution f(x) = 2x5 7x4 + 12x3 + 2x2 + 4x + 6 The coefficients in f(x) have _2_ sign changes, so f has __2 or 0__ positive real zero(s). f(x) = 2(x)5 7(x)4 + 12(x)3 + 2(x)2 + 4(x) + 6 f(x) = _2x5 7x4 12x3 + 2x2 4x + 6_ The coefficients in f(x) have _3_ sign changes, so f has _3 or 1_ negative real zero(s). Positive real zeros Negative real zeros Imaginary zeros Total zeros _2_ _3_ _0_ _5_ _2_ _1_ _2_ _5_ _0_ _3_ _2_ _5_ _0_ _1_ _4_ _5_ Checkpoint Complete the following exercise. 4. Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for f(x) = 3x5 4x4 + x3 + 6x2 + 7x 8. Positive Negative real zeros real zeros Imaginary zeros Total zeros 3 2 0 5 3 0 2 5 1 2 2 5 1 0 4 5 Homework _______________________________________________________________________ _______________________________________________________________________