To determine which of the given values are roots of the polynomial [tex]\( F(x) = x^3 - 7x^2 + 13x - 6 \)[/tex], we need to substitute each value into the polynomial and check if it equals zero. A value [tex]\( a \)[/tex] is a root if [tex]\( F(a) = 0 \)[/tex].
Step-by-Step Verification:
1. Option A: [tex]\( \frac{5 + \sqrt{13}}{2} \)[/tex]
Substitute [tex]\( x = \frac{5 + \sqrt{13}}{2} \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[
F\left(\frac{5 + \sqrt{13}}{2}\right) \neq 0
\][/tex]
2. Option B: [tex]\( -3 \)[/tex]
Substitute [tex]\( x = -3 \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[
F(-3) \neq 0
\][/tex]
3. Option C: [tex]\( 2 \)[/tex]
Substitute [tex]\( x = 2 \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[
F(2) = 2^3 - 7 \cdot 2^2 + 13 \cdot 2 - 6
\][/tex]
Simplify to check:
[tex]\[
F(2) = 8 - 28 + 26 - 6 = 0
\][/tex]
Since [tex]\( F(2) = 0 \)[/tex], [tex]\( 2 \)[/tex] is a root.
4. Option D: [tex]\( \frac{5 - \sqrt{13}}{2} \)[/tex]
Substitute [tex]\( x = \frac{5 - \sqrt{13}}{2} \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[
F\left(\frac{5 - \sqrt{13}}{2}\right) = 0
\][/tex]
Since [tex]\( F\left(\frac{5 - \sqrt{13}}{2}\right) = 0 \)[/tex], [tex]\( \frac{5 - \sqrt{13}}{2} \)[/tex] is a root.
5. Option E: [tex]\( \frac{3 + \sqrt{12}}{4} \)[/tex]
Substitute [tex]\( x = \frac{3 + \sqrt{12}}{4} \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[
F\left(\frac{3 + \sqrt{12}}{4}\right) \neq 0
\][/tex]
6. Option F: [tex]\( \frac{3 - \sqrt{12}}{4} \)[/tex]
Substitute [tex]\( x = \frac{3 - \sqrt{12}}{4} \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[
F\left(\frac{3 - \sqrt{12}}{4}\right) \neq 0
\][/tex]
So, the values that are roots of the polynomial [tex]\( F(x) = x^3 - 7x^2 + 13x - 6 \)[/tex] are:
- [tex]\( 2 \)[/tex]
- [tex]\( \frac{5 - \sqrt{13}}{2} \)[/tex]
Thus, the correct answers are:
- C. [tex]\( 2 \)[/tex]
- D. [tex]\( \frac{5 - \sqrt{13}}{2} \)[/tex]
