To find the full solution set for the equation \( x^3 + 5x^2 + 10x + 50 = 0 \) given that one solution is \( x = -5 \), we can use polynomial long division to factor the cubic polynomial.
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Use polynomial long division to divide \( x^3 + 5x^2 + 10x + 50 \) by \( x + 5 \):
\[ x^3 + 5x^2 + 10x + 50 \div (x + 5) \]
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The first term is \( x^2 \). Multiply \( x + 5 \) by \( x^2 \): \[ x^3 + 5x^2 \] Subtract: \[ (x^3 + 5x^2 + 10x + 50) - (x^3 + 5x^2) = 10x + 50 \]
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The second term is \( 10 \). Multiply \( x + 5 \) by \( 10 \): \[ 10x + 50 \] Subtract: \[ (10x + 50) - (10x + 50) = 0 \]
Therefore, we have: \[ x^3 + 5x^2 + 10x + 50 = (x + 5)(x^2 + 10) \]
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Find the solutions of the quadratic \( x^2 + 10 = 0 \):
Solving for \( x \): \[ x^2 = -10 \] \[ x = \pm i\sqrt{10} \]
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The full solution set for the original cubic equation is:
- One real solution: \( -5 \)
- Two non-real solutions: \( i\sqrt{10} \) and \( -i\sqrt{10} \)
Based on the analysis above, the solution set consists of:
- One real solution and two non-real solutions.
Therefore, the correct option is: Option #1: It has one real solution and two non-real solutions.