If $\alpha, \beta$ are the roots of the equation $x^2+3x+2=0$. Then $\alpha^5+\beta^5=?$

Using Vieta's formulas for the equation x^2 + 3x + 2 = 0, the sum of the roots alpha + beta equals -3 and the product alpha * beta equals 2. We can square the sum of the roots to get alpha^2 + beta^2 + 2(alpha)(beta) = (-3)^2, which simplifies to alpha^2 + beta^2 = 9 - 4 = 5. Next, we multiply alpha^2 + beta^2 = 5 by the sum alpha + beta = -3 to obtain alpha^3 + beta^3 + (alpha)(beta)(alpha + beta) = -15. Substituting the known values gives alpha^3 + beta^3 + 2(-3) = -15, resulting in alpha^3 + beta^3 = -9. Finally, we multiply alpha^3 + beta^3 = -9 by the sum alpha + beta = -3 to get alpha^4 + beta^4 + (alpha)(beta)(alpha^2 + beta^2) = 27, which yields alpha^4 + beta^4 = 17. We then multiply alpha^4 + beta^4 = 17 by the sum alpha + beta = -3 to obtain alpha^5 + beta^5 + (alpha)(beta)(alpha^3 + beta^3) = -51. Substituting the known values of alpha * beta = 2 and alpha^3 + beta^3 = -9 gives alpha^5 + beta^5 + 2(-9) = -51. Solving this yields alpha^5 + beta^5 = -33.

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