Answer :
To find a polynomial function of degree 3 with given zeros [tex]\(5 - \sqrt{2}, 5 + \sqrt{2},\)[/tex] and [tex]\(-10\)[/tex], we follow these steps:
### Step 1: Use the fact that the polynomial can be written in terms of its zeros
If a polynomial has zeros [tex]\(r_1, r_2, \ldots, r_n\)[/tex], it can be expressed as:
[tex]\[ P(x) = a(x - r_1)(x - r_2) \ldots (x - r_n) \][/tex]
Given that the leading coefficient is 1, our polynomial function will be:
[tex]\[ P(x) = (x - (5 - \sqrt{2}))(x - (5 + \sqrt{2}))(x - (-10)) \][/tex]
[tex]\[ P(x) = (x - 5 + \sqrt{2})(x - 5 - \sqrt{2})(x + 10) \][/tex]
### Step 2: Simplify the polynomial by multiplying the factors
First, we will simplify the product of the first two factors. Notice that these factors are of the form [tex]\((a - b)(a + b)\)[/tex], which is a difference of squares:
[tex]\[ (x - 5 + \sqrt{2})(x - 5 - \sqrt{2}) = ((x - 5) + \sqrt{2})((x - 5) - \sqrt{2}) \][/tex]
[tex]\[ = (x - 5)^2 - (\sqrt{2})^2 \][/tex]
[tex]\[ = (x - 5)^2 - 2 \][/tex]
Next, we expand [tex]\((x - 5)^2\)[/tex]:
[tex]\[ (x - 5)^2 = x^2 - 2 \cdot 5 \cdot x + 5^2 \][/tex]
[tex]\[ = x^2 - 10x + 25 \][/tex]
Now, substitute this back into the expression:
[tex]\[ (x - 5)^2 - 2 = x^2 - 10x + 25 - 2 \][/tex]
[tex]\[ = x^2 - 10x + 23 \][/tex]
Now, we have:
[tex]\[ P(x) = (x^2 - 10x + 23)(x + 10) \][/tex]
### Step 3: Expand the simplified product
[tex]\[ P(x) = (x^2 - 10x + 23)(x + 10) \][/tex]
We distribute each term in [tex]\((x + 10)\)[/tex] through the quadratic polynomial [tex]\(x^2 - 10x + 23\)[/tex]:
[tex]\[ P(x) = x^2(x + 10) + (-10x)(x + 10) + 23(x + 10) \][/tex]
Simplify each product:
[tex]\[ P(x) = x^3 + 10x^2 - 10x^2 - 100x + 23x + 230 \][/tex]
Combine like terms:
[tex]\[ P(x) = x^3 + (10x^2 - 10x^2) + (-100x + 23x) + 230 \][/tex]
[tex]\[ P(x) = x^3 - 77x + 230 \][/tex]
### Final Polynomial
[tex]\[ P(x) = x^3 - 77x + 230 \][/tex]
Thus, the polynomial function of degree 3 with the given zeros [tex]\(5 - \sqrt{2}, 5 + \sqrt{2},\)[/tex] and [tex]\(-10\)[/tex] is:
[tex]\[ P(x) = x^3 - 77x + 230 \][/tex]
### Step 1: Use the fact that the polynomial can be written in terms of its zeros
If a polynomial has zeros [tex]\(r_1, r_2, \ldots, r_n\)[/tex], it can be expressed as:
[tex]\[ P(x) = a(x - r_1)(x - r_2) \ldots (x - r_n) \][/tex]
Given that the leading coefficient is 1, our polynomial function will be:
[tex]\[ P(x) = (x - (5 - \sqrt{2}))(x - (5 + \sqrt{2}))(x - (-10)) \][/tex]
[tex]\[ P(x) = (x - 5 + \sqrt{2})(x - 5 - \sqrt{2})(x + 10) \][/tex]
### Step 2: Simplify the polynomial by multiplying the factors
First, we will simplify the product of the first two factors. Notice that these factors are of the form [tex]\((a - b)(a + b)\)[/tex], which is a difference of squares:
[tex]\[ (x - 5 + \sqrt{2})(x - 5 - \sqrt{2}) = ((x - 5) + \sqrt{2})((x - 5) - \sqrt{2}) \][/tex]
[tex]\[ = (x - 5)^2 - (\sqrt{2})^2 \][/tex]
[tex]\[ = (x - 5)^2 - 2 \][/tex]
Next, we expand [tex]\((x - 5)^2\)[/tex]:
[tex]\[ (x - 5)^2 = x^2 - 2 \cdot 5 \cdot x + 5^2 \][/tex]
[tex]\[ = x^2 - 10x + 25 \][/tex]
Now, substitute this back into the expression:
[tex]\[ (x - 5)^2 - 2 = x^2 - 10x + 25 - 2 \][/tex]
[tex]\[ = x^2 - 10x + 23 \][/tex]
Now, we have:
[tex]\[ P(x) = (x^2 - 10x + 23)(x + 10) \][/tex]
### Step 3: Expand the simplified product
[tex]\[ P(x) = (x^2 - 10x + 23)(x + 10) \][/tex]
We distribute each term in [tex]\((x + 10)\)[/tex] through the quadratic polynomial [tex]\(x^2 - 10x + 23\)[/tex]:
[tex]\[ P(x) = x^2(x + 10) + (-10x)(x + 10) + 23(x + 10) \][/tex]
Simplify each product:
[tex]\[ P(x) = x^3 + 10x^2 - 10x^2 - 100x + 23x + 230 \][/tex]
Combine like terms:
[tex]\[ P(x) = x^3 + (10x^2 - 10x^2) + (-100x + 23x) + 230 \][/tex]
[tex]\[ P(x) = x^3 - 77x + 230 \][/tex]
### Final Polynomial
[tex]\[ P(x) = x^3 - 77x + 230 \][/tex]
Thus, the polynomial function of degree 3 with the given zeros [tex]\(5 - \sqrt{2}, 5 + \sqrt{2},\)[/tex] and [tex]\(-10\)[/tex] is:
[tex]\[ P(x) = x^3 - 77x + 230 \][/tex]