Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1.

Given zeros: [tex]\(-1, 3, 6\)[/tex]

The polynomial function is [tex]\( f(x) = \square \)[/tex].

(Simplify your answer. Use integers or fractions for any numbers in the expression.)



Answer :

To find a polynomial function of degree 3 with the given numbers as its zeros, we need to form a polynomial whose roots are [tex]\(-1\)[/tex], [tex]\(3\)[/tex], and [tex]\(6\)[/tex]. The general form of a polynomial with these roots can be written as:

[tex]\[ f(x) = (x - \text{root}_1)(x - \text{root}_2)(x - \text{root}_3) \][/tex]

Here, the roots are given as [tex]\(-1\)[/tex], [tex]\(3\)[/tex], and [tex]\(6\)[/tex]. So, the polynomial function [tex]\(f(x)\)[/tex] can be written as:

[tex]\[ f(x) = (x - (-1))(x - 3)(x - 6) \][/tex]

Simplifying the terms inside the parentheses:

[tex]\[ f(x) = (x + 1)(x - 3)(x - 6) \][/tex]

Next, we will expand this expression step by step:

First, expand [tex]\((x + 1)(x - 3)\)[/tex]:

[tex]\[ (x + 1)(x - 3) = x(x - 3) + 1(x - 3) \][/tex]

[tex]\[ = x^2 - 3x + x - 3 \][/tex]

[tex]\[ = x^2 - 2x - 3 \][/tex]

Now we need to multiply this result by [tex]\((x - 6)\)[/tex]:

[tex]\[ (x^2 - 2x - 3)(x - 6) \][/tex]

Distribute each term in the first polynomial by each term in the second polynomial:

[tex]\[ = x^2(x - 6) - 2x(x - 6) - 3(x - 6) \][/tex]

[tex]\[ = x^3 - 6x^2 - 2x^2 + 12x - 3x + 18 \][/tex]

Combine like terms:

[tex]\[ = x^3 - 8x^2 + 9x + 18 \][/tex]

Thus, the polynomial function of degree 3 with the zeros [tex]\(-1\)[/tex], [tex]\(3\)[/tex], and [tex]\(6\)[/tex] is:

[tex]\[ f(x) = x^3 - 8x^2 + 9x + 18 \][/tex]