Answer :
To solve for [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex], where [tex]\(\alpha, \beta, \gamma\)[/tex] are the roots of the polynomial equation [tex]\(x^3 + 2x + 5 = 0\)[/tex], let's start by using the relationships provided by Vieta's formulas for a cubic polynomial.
Given the polynomial [tex]\(x^3 + 2x + 5 = 0\)[/tex], we can identify the coefficients as follows: [tex]\(a = 1\)[/tex], [tex]\(b = 0\)[/tex], [tex]\(c = 2\)[/tex], and [tex]\(d = 5\)[/tex]. Therefore, according to Vieta’s formulas, we have:
1. [tex]\(\alpha + \beta + \gamma = 0\)[/tex],
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)[/tex],
3. [tex]\(\alpha\beta\gamma = -5\)[/tex].
Next, we need to express [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] in terms of these roots. Let's begin by finding the squared roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex]. Using the original equation [tex]\(x^3 + 2x + 5 = 0\)[/tex]:
[tex]\[ \alpha^3 + 2\alpha + 5 = 0 \implies \alpha^3 = -2\alpha - 5, \][/tex]
[tex]\[ \beta^3 + 2\beta + 5 = 0 \implies \beta^3 = -2\beta - 5, \][/tex]
[tex]\[ \gamma^3 + 2\gamma + 5 = 0 \implies \gamma^3 = -2\gamma - 5. \][/tex]
Now, squaring [tex]\(\alpha^2\)[/tex], we have:
[tex]\[ \alpha^4 = (\alpha^2)^2 = (\alpha^3 \cdot \alpha) = (-2\alpha - 5)\alpha = -2\alpha^2 - 5\alpha, \][/tex]
[tex]\[ \beta^4 = (\beta^2)^2 = (\beta^3 \cdot \beta) = (-2\beta - 5)\beta = -2\beta^2 - 5\beta, \][/tex]
[tex]\[ \gamma^4 = (\gamma^2)^2 = (\gamma^3 \cdot \gamma) = (-2\gamma - 5)\gamma = -2\gamma^2 - 5\gamma. \][/tex]
Summing these expressions, we get:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2) - 5(\alpha + \beta + \gamma). \][/tex]
Since [tex]\(\alpha + \beta + \gamma = 0\)[/tex], the term involving [tex]\(\alpha + \beta + \gamma\)[/tex] vanishes:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2). \][/tex]
To find [tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex], recall the identity for the sum of squares of the roots:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha). \][/tex]
Using Vieta’s results:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = 0^2 - 2(2) = -4. \][/tex]
Thus,
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(-4) = 8. \][/tex]
Therefore, the value of [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] is [tex]\(\boxed{8}\)[/tex].
Given the polynomial [tex]\(x^3 + 2x + 5 = 0\)[/tex], we can identify the coefficients as follows: [tex]\(a = 1\)[/tex], [tex]\(b = 0\)[/tex], [tex]\(c = 2\)[/tex], and [tex]\(d = 5\)[/tex]. Therefore, according to Vieta’s formulas, we have:
1. [tex]\(\alpha + \beta + \gamma = 0\)[/tex],
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)[/tex],
3. [tex]\(\alpha\beta\gamma = -5\)[/tex].
Next, we need to express [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] in terms of these roots. Let's begin by finding the squared roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex]. Using the original equation [tex]\(x^3 + 2x + 5 = 0\)[/tex]:
[tex]\[ \alpha^3 + 2\alpha + 5 = 0 \implies \alpha^3 = -2\alpha - 5, \][/tex]
[tex]\[ \beta^3 + 2\beta + 5 = 0 \implies \beta^3 = -2\beta - 5, \][/tex]
[tex]\[ \gamma^3 + 2\gamma + 5 = 0 \implies \gamma^3 = -2\gamma - 5. \][/tex]
Now, squaring [tex]\(\alpha^2\)[/tex], we have:
[tex]\[ \alpha^4 = (\alpha^2)^2 = (\alpha^3 \cdot \alpha) = (-2\alpha - 5)\alpha = -2\alpha^2 - 5\alpha, \][/tex]
[tex]\[ \beta^4 = (\beta^2)^2 = (\beta^3 \cdot \beta) = (-2\beta - 5)\beta = -2\beta^2 - 5\beta, \][/tex]
[tex]\[ \gamma^4 = (\gamma^2)^2 = (\gamma^3 \cdot \gamma) = (-2\gamma - 5)\gamma = -2\gamma^2 - 5\gamma. \][/tex]
Summing these expressions, we get:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2) - 5(\alpha + \beta + \gamma). \][/tex]
Since [tex]\(\alpha + \beta + \gamma = 0\)[/tex], the term involving [tex]\(\alpha + \beta + \gamma\)[/tex] vanishes:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2). \][/tex]
To find [tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex], recall the identity for the sum of squares of the roots:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha). \][/tex]
Using Vieta’s results:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = 0^2 - 2(2) = -4. \][/tex]
Thus,
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(-4) = 8. \][/tex]
Therefore, the value of [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] is [tex]\(\boxed{8}\)[/tex].
