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What is the polynomial function of the lowest degree with a leading coefficient of 1 and roots [tex]\sqrt{3}, -4[/tex], and [tex]4[/tex]?

A. [tex]f(x) = x^3 - 3x^2 + 16x + 48[/tex]

B. [tex]f(x) = x^3 - 3x^2 - 16x + 48[/tex]

C. [tex]f(x) = x^4 - 19x^2 + 48[/tex]

D. [tex]f(x) = x^4 - 13x^2 + 48[/tex]

Answer :

To find the polynomial of lowest degree with leading coefficient 1 and with the given roots, we proceed with the following steps:

1. Since one of the roots is [tex]$ \sqrt{3} $[/tex], and we want the polynomial to have integer coefficients, its conjugate [tex]$ -\sqrt{3} $[/tex] must also be a root.

2. The given roots, therefore, are:
[tex]$$ \sqrt{3}, \quad -\sqrt{3}, \quad -4, \quad 4. $$[/tex]

3. The factors corresponding to these roots are:
- For [tex]$ \sqrt{3} $[/tex] and [tex]$ -\sqrt{3} $[/tex], the factors are:
[tex]$$ (x - \sqrt{3})(x + \sqrt{3}) = x^2 - 3. $$[/tex]
- For [tex]$ -4 $[/tex] and [tex]$ 4 $[/tex], the factors are:
[tex]$$ (x - 4)(x + 4) = x^2 - 16. $$[/tex]

4. The polynomial is the product of these factors:
[tex]$$ f(x) = (x^2 - 3)(x^2 - 16). $$[/tex]

5. Expanding the product:
[tex]\[
\begin{aligned}
f(x) &= x^2(x^2 - 16) - 3(x^2 - 16) \\
&= x^4 - 16x^2 - 3x^2 + 48 \\
&= x^4 - 19x^2 + 48.
\end{aligned}
\][/tex]

Thus, the polynomial function of lowest degree with leading coefficient 1 and the given roots is

[tex]$$ f(x) = x^4 - 19x^2 + 48. $$[/tex]

Among the given options, this corresponds to option 3.

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