To find a polynomial function of degree 3 with given numbers as zeros, we'll follow these steps:
1. Identify the zeros: The given zeros are [tex]\(3 - \sqrt{3}\)[/tex], [tex]\(3 + \sqrt{3}\)[/tex], and [tex]\(-6\)[/tex].
2. Form the factors corresponding to each zero:
For a zero [tex]\(a\)[/tex], the corresponding factor is [tex]\((x - a)\)[/tex].
- For the zero [tex]\(3 - \sqrt{3}\)[/tex], the factor is [tex]\((x - (3 - \sqrt{3})) = (x - 3 + \sqrt{3})\)[/tex].
- For the zero [tex]\(3 + \sqrt{3}\)[/tex], the factor is [tex]\((x - (3 + \sqrt{3})) = (x - 3 - \sqrt{3})\)[/tex].
- For the zero [tex]\(-6\)[/tex], the factor is [tex]\((x - (-6)) = (x + 6)\)[/tex].
3. Set up the polynomial as the product of these factors:
[tex]\[
f(x) = (x - (3 - \sqrt{3}))(x - (3 + \sqrt{3}))(x + 6)
\][/tex]
4. Simplify the product:
- First, simplify the product [tex]\((x - (3 - \sqrt{3}))(x - (3 + \sqrt{3}))\)[/tex]:
Notice that this product is in the form [tex]\((a - b)(a + b)\)[/tex], which simplifies to [tex]\(a^2 - b^2\)[/tex].
Here, [tex]\(a = x - 3\)[/tex] and [tex]\(b = \sqrt{3}\)[/tex].
So,
[tex]\[
(x - 3 + \sqrt{3})(x - 3 - \sqrt{3}) = ((x - 3)^2 - (\sqrt{3})^2)
\][/tex]
[tex]\[
((x - 3)^2 - 3)
\][/tex]
Expand [tex]\((x - 3)^2\)[/tex]:
[tex]\[
(x - 3)^2 = x^2 - 6x + 9
\][/tex]
So,
[tex]\[
(x^2 - 6x + 9) - 3 = x^2 - 6x + 6
\][/tex]
- Now, we need to multiply this result by the third factor [tex]\((x + 6)\)[/tex]:
[tex]\[
f(x) = (x^2 - 6x + 6)(x + 6)
\][/tex]
5. Distribute to find the expanded polynomial:
[tex]\[
f(x) = x^2(x + 6) - 6x(x + 6) + 6(x + 6)
\][/tex]
Distribute each term:
[tex]\[
f(x) = (x^3 + 6x^2) - (6x^2 + 36x) + (6x + 36)
\][/tex]
Combine like terms:
[tex]\[
f(x) = x^3 + 6x^2 - 6x^2 - 36x + 6x + 36
\][/tex]
[tex]\[
f(x) = x^3 - 30x + 36
\][/tex]
So, the polynomial function is:
[tex]\[
f(x) = x^3 - 30x + 36
\][/tex]
