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To solve the equation [tex]\(x^3 + 3x^2 - 5x - 15 = 0\)[/tex] for [tex]\(x\)[/tex] over the set of real numbers, we follow a structured approach to find the roots.
1. Identify the problem: We need to find the real roots of the cubic equation [tex]\(x^3 + 3x^2 - 5x - 15 = 0\)[/tex].
2. Analyze possible roots: We start by considering possible rational roots using the Rational Root Theorem, which states that any rational root of the polynomial equation, where the polynomial is in standard form [tex]\( ax^n + bx^{n-1} + ... + k = 0 \)[/tex], must be a factor of the constant term ([tex]\(-15\)[/tex]) divided by the leading coefficient (which is 1 here). So we look at the factors of [tex]\(-15\)[/tex]: [tex]\( \pm 1, \pm 3, \pm 5, \pm 15 \)[/tex].
3. Testing possible roots:
- For [tex]\(x = -3\)[/tex]:
[tex]\[ (-3)^3 + 3(-3)^2 - 5(-3) - 15 = -27 + 27 + 15 - 15 = 0 \][/tex]
Hence, [tex]\(x = -3\)[/tex] is a root.
4. Factor the polynomial using [tex]\(x = -3\)[/tex]:
We can then factor out [tex]\(x + 3\)[/tex] from the polynomial. Using polynomial division or synthetic division, we divide [tex]\(x^3 + 3x^2 - 5x - 15\)[/tex] by [tex]\(x + 3\)[/tex]:
[tex]\[ (x^3 + 3x^2 - 5x - 15) \div (x + 3) = x^2 - 5 \][/tex]
Thus, we have:
[tex]\[ x^3 + 3x^2 - 5x - 15 = (x + 3)(x^2 - 5) \][/tex]
5. Solve the quadratic equation [tex]\( x^2 - 5 = 0 \)[/tex]:
- Set [tex]\( x^2 - 5 = 0 \)[/tex]:
[tex]\[ x^2 - 5 = 0 \][/tex]
[tex]\[ x^2 = 5 \][/tex]
[tex]\[ x = \pm\sqrt{5} \][/tex]
6. Combine all solutions:
From the steps above, we have found that the solutions to the equation [tex]\(x^3 + 3x^2 - 5x - 15 = 0\)[/tex] are:
[tex]\[ x = -3, \sqrt{5}, -\sqrt{5} \][/tex]
Therefore, the exact solutions over the set of real numbers are:
[tex]\[ x = -3, \sqrt{5}, -\sqrt{5} \][/tex]
1. Identify the problem: We need to find the real roots of the cubic equation [tex]\(x^3 + 3x^2 - 5x - 15 = 0\)[/tex].
2. Analyze possible roots: We start by considering possible rational roots using the Rational Root Theorem, which states that any rational root of the polynomial equation, where the polynomial is in standard form [tex]\( ax^n + bx^{n-1} + ... + k = 0 \)[/tex], must be a factor of the constant term ([tex]\(-15\)[/tex]) divided by the leading coefficient (which is 1 here). So we look at the factors of [tex]\(-15\)[/tex]: [tex]\( \pm 1, \pm 3, \pm 5, \pm 15 \)[/tex].
3. Testing possible roots:
- For [tex]\(x = -3\)[/tex]:
[tex]\[ (-3)^3 + 3(-3)^2 - 5(-3) - 15 = -27 + 27 + 15 - 15 = 0 \][/tex]
Hence, [tex]\(x = -3\)[/tex] is a root.
4. Factor the polynomial using [tex]\(x = -3\)[/tex]:
We can then factor out [tex]\(x + 3\)[/tex] from the polynomial. Using polynomial division or synthetic division, we divide [tex]\(x^3 + 3x^2 - 5x - 15\)[/tex] by [tex]\(x + 3\)[/tex]:
[tex]\[ (x^3 + 3x^2 - 5x - 15) \div (x + 3) = x^2 - 5 \][/tex]
Thus, we have:
[tex]\[ x^3 + 3x^2 - 5x - 15 = (x + 3)(x^2 - 5) \][/tex]
5. Solve the quadratic equation [tex]\( x^2 - 5 = 0 \)[/tex]:
- Set [tex]\( x^2 - 5 = 0 \)[/tex]:
[tex]\[ x^2 - 5 = 0 \][/tex]
[tex]\[ x^2 = 5 \][/tex]
[tex]\[ x = \pm\sqrt{5} \][/tex]
6. Combine all solutions:
From the steps above, we have found that the solutions to the equation [tex]\(x^3 + 3x^2 - 5x - 15 = 0\)[/tex] are:
[tex]\[ x = -3, \sqrt{5}, -\sqrt{5} \][/tex]
Therefore, the exact solutions over the set of real numbers are:
[tex]\[ x = -3, \sqrt{5}, -\sqrt{5} \][/tex]
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