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Answer :
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!
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!
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