4Two Major TheoremsFundamental Theorem of Algebra: A polynomial function of degree n has n complex zeros( real or nonreal). Some of these zeros may be repeated.Liner Factorization Theorem: If f(x) is a polynomial function of degree n >0, then f(x) has precisely n linear factors and f(x) = a(x –z1)(x –z2) (x -zn)Where a is the leading coefficient of f(x) and z1, z2,….., zn are the complex zeros of f(x). The zi are not necessarily distinct numbers; some may be repeated.
5Fundamental Polynomial Connections in the Complex Case: The following statements about a polynomial function f are equivalent if k is a complex number. 1. x = k is a solution (or root) of the equation f(x) = 0 2. is a zero of the function f. 3.x - k is a factor of f(x).
6Complex Conjugate Zeros Suppose that f(x) is a polynomial function with real coefficients. If a and b are real numbers with b not equal 0 and a + bi is a zero of f(x), then its complex conjugate a - bi is also a zero of f(x).
7Factoring with Real Number Coefficients Let f(s) be a polynomial function with real coefficients. The Linear Factorization Theorem tells us that f(x) can be factored into the form f(x) = a(x –z1)(x –z2) (x -zn), Where zi are complex numbers. Recall, however, that nonreal complex zeros occur in conjugate pairs. The product of x - (a + bi) and x - (a - bi) is [x-(a+bi)][x-(a-bi)] = x^(2) - (a-bi)x - (a+bi)x + (a+bi)(a-bi) = x^(2 ) - 2ax + ((a^2) + (b^2)).
8Polynomial Function of Odd Degree Factoring with Real Number Coefficients Every polynomial function with real coefficients can be written as a product of linear factors and irreducible quadratic factors, each with real coefficients.Polynomial Function of Odd DegreeEvery polynomial function of odd degree with real coefficients has at least one real zero.
9Complex Zeros of Polynomial Find complex zeros of polynomial function :F(x) = 3x^(4) + 5x^(3) + 25x^(2) + 45x – 18
11Assessment of ContentWrite the polynomial in standard from, and identify the zeros of the function and the x-intercept of it’s graph.1.f(x)=(x-3i)(x+3i)A. x2+9; zeros:+-3i: x-intercepts: noneB. x2+6; zeros:+-3i: x-intercepts: 22.f(x)=(x-1)(x-1)(x+2i)(x-2i)A. X4-3x+25x-4x+4B. X4-2x+5x-8x+4Write the polynomial function of minimum degree in standard form with real coefficient whose zeros included those listed.3. I and – IA. x3+4B. x2+14. 2,3 and IA. X4-5x3+7x-5x+6B. X2-2x2+7x-3x+4State how many complex and real zeros the function has.5. f(x)=x2-2x+7A. 2 complex zeros: none realB. 4 complex zeros: 7 real6. F(x)=x4-5x3+x2-3x+6
12Assessment of Content continue Find all of the zeros and write a linear factorization of the function.7. f(x)= x3+4x-5A. Zeros: x=1, x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(2x+1+ √19i) (2x+1- √ 19i)B. Zeros: x=2, x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(2x+1+ √19i) (2x+1- √ 19i)8. f(x)= x4+x3+5x2-x-6A. Zeros: x= ± 1, x = -1/2 ± √ 23/2i ; f(x) = ¼ (x-1)(x+1)(2x+1+ √ 19i) (2x+1- √ 23iB. . Zeros: x= ± 1, x = -1/2 ± √ 23/2i ; f(x) = ¼ (x-1)(x+1)(2x+1+ √ 19i) (2x+1- √ 23iUsing the given zeros, find all of the zeros and write a linear factorization of f(x).9. 1+i is a zeros of f(x)=x4-2x3-x2+6x-6A. Zeros: x=2 x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(3x+1+ √ 19i) (3x+1- √ 19i)B. Zeros: x=1, x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(2x+1+ √ 19i) (2x+1- √ 19i)i is a zeros of f(x)=x4-6x3+11x2+12x-26A. Zeros: x +- √ 2. x =3 ± 2i: f(x) = (x- √ 2) ( x+ √ 2)(x-3+2i)(x-3-2i).B. Zeros: x +- √ 4. x =3 ± 3i: f(x) = (x- √ 2) ( x+ √ 2)(x-3+2i)(x-3-2i).