Finding the Zeros of a Polynomial Function

In Exercises 59-70, write the polynomial as the product of linear factors and list all the zeros of the function.

59. [tex]f(x) = x^2 + 36[/tex]

60. [tex]f(x) = x^2 + 49[/tex]

61. [tex]h(x) = x^2 - 2x + 17[/tex]

62. [tex]g(x) = x^2 + 10x + 17[/tex]

63. [tex]f(x) = x^4 - 16[/tex]

64. [tex]f(y) = y^4 - 256[/tex]

65. [tex]f(z) = z^2 - 2z + 2[/tex]

66. [tex]h(x) = x^3 - 3x^2 + 4x - 2[/tex]

67. [tex]g(x) = x^3 - 3x^2 + x + 5[/tex]

68. [tex]f(x) = x^3 - x^2 + x + 39[/tex]

69. [tex]g(x) = x^4 - 4x^3 + 8x^2 - 16x + 16[/tex]

70. [tex]h(x) = x^4 + 6x^3 + 10x^2 + 6x + 9[/tex]

To find the zeros of a polynomial function and write it as the product of linear factors, let's start with the first polynomial given in the exercises:

59. [tex]\( f(x) = x^2 + 36 \)[/tex]

Step 1: Factor the polynomial.
This polynomial can be rewritten using the identity for the sum of squares:

[tex]\[ f(x) = x^2 + 36 = (x + 6i)(x - 6i) \][/tex]

Here, [tex]\(i\)[/tex] is the imaginary unit, where [tex]\(i = \sqrt{-1}\)[/tex].

Step 2: Find the zeros of the polynomial.
To find the zeros, set each factor equal to zero:

[tex]\[ x + 6i = 0 \quad \Rightarrow \quad x = -6i \][/tex]

[tex]\[ x - 6i = 0 \quad \Rightarrow \quad x = 6i \][/tex]

So, the zeros of the function are [tex]\( x = -6i \)[/tex] and [tex]\( x = 6i \)[/tex].

In summary:
- The polynomial [tex]\( f(x) = x^2 + 36 \)[/tex] is factored as [tex]\((x + 6i)(x - 6i)\)[/tex].
- The zeros of the polynomial are [tex]\( x = -6i \)[/tex] and [tex]\( x = 6i \)[/tex].

You can apply similar methods to other polynomials in the exercise list. Let me know if you want a detailed solution for another item on the list!