First, recall that \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2 + 5x + 6 = 0\). By Vieta's formulas, the sum and product of the roots for a quadratic equation \(ax^2 + bx + c = 0\) are given by:
[tex]\[
\alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a}
\][/tex]
For our specific equation \(x^2 + 5x + 6 = 0\):
[tex]\[
a = 1, \, b = 5, \, c = 6
\][/tex]
Thus,
[tex]\[
\alpha + \beta = -\frac{5}{1} = -5
\][/tex]
[tex]\[
\alpha \beta = \frac{6}{1} = 6
\][/tex]
Now, we need to find \(\alpha^5 + \beta^5\). We can relate higher powers of \(\alpha\) and \(\beta\) using their lower powers. Using the identity:
[tex]\[
\alpha^5 + \beta^5 = (\alpha + \beta)(\alpha^4 + \beta^4) - \alpha \beta (\alpha^3 + \beta^3)
\][/tex]
We need to express \(\alpha^4 + \beta^4\) and \(\alpha^3 + \beta^3\) in terms of \(\alpha + \beta\) and \(\alpha \beta\).
First, for \(\alpha^3 + \beta^3\), use the identity:
[tex]\[
\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha \beta)
\][/tex]
To find \(\alpha^2 + \beta^2\), use the identity:
[tex]\[
\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta
\][/tex]
Substitute the known values:
[tex]\[
\alpha^2 + \beta^2 = (-5)^2 - 2 \cdot 6 = 25 - 12 = 13
\][/tex]
Thus,
[tex]\[
\alpha^3 + \beta^3 = (-5)(13 - 6) = -5 \cdot 7 = -35
\][/tex]
Next, for \(\alpha^4 + \beta^4\), use the identity:
[tex]\[
\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2
\][/tex]
Substitute the known values:
[tex]\[
\alpha^4 + \beta^4 = (13)^2 - 2 \cdot 6^2 = 169 - 2 \cdot 36 = 169 - 72 = 97
\][/tex]
Now, substitute everything back into the identity for \(\alpha^5 + \beta^5\):
[tex]\[
\alpha^5 + \beta^5 = (\alpha + \beta)(\alpha^4 + \beta^4) - \alpha \beta (\alpha^3 + \beta^3)
\][/tex]
Substitute the known values:
[tex]\[
\alpha^5 + \beta^5 = (-5)(97) - 6(-35)
\][/tex]
[tex]\[
\alpha^5 + \beta^5 = -485 + 210 = -275
\][/tex]
Therefore, the value of \(\alpha^5 + \beta^5\) is:
[tex]\[
\boxed{-275}
\][/tex]
