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Do Now: Find all real zeros of the function. f ( x) 2 x 5x 18x 19 x 42 4 3 2 Academy Algebra II 5.7: Apply the Fundamental Theorem of Algebra HW Friday: p.384 (12-18 even) Test 5.5-5.9: Tuesday, 12/2 Descartes’ Rule of Signs • The number of positive real zeros of f is equal to the number of sign changes in the 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. Fundamental Theorem of Algebra • If f(x) is a polynomial with degree of n (where n>0), then the equation f(x) = 0 has at least one solution. • Corollary: 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. Fundamental Theorem of Algebra • Example: x3 – 5x2 – 8x + 48 = 0 (x+3)(x – 4)2= 0 x = -3 and x = 4 This equation only has two distinct solutions: -3 and 4. Because the factor (x – 4) appears twice, you can count the solution 4 twice. With the repeated solution the third-degree equation has three solutions. Solve the equation. x3 + 5x2 + 4x + 20 = 0 Find the zeros of the polynomial function. f(x) = x4 – 8x3 + 18x2 – 27 Find all zeros of the polynomial function. f(x) = x5 – 2x4 + 8x2 – 13x + 6 Conjugates Theorem • If a bi is an imaginary zero of function f, then a bi is also a zero. • If a b is a zero of function f where is irrational, then a b is also a zero. b Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, & the given zeros. 1.) -1, 2, 4 2.) 4, 1 5 Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, & the given zeros. 3.) 2, 2i, 4 6