Answer :
To determine the correct statement regarding the roots of the polynomial equation \( x^5 = -2x^2 \), let's analyze the equation step by step.
1. Rewrite the Equation: Start by rewriting the equation in standard polynomial form:
[tex]\[ x^5 + 2x^2 = 0 \][/tex]
2. Factor Out the Common Term: Factor out the common term \( x^2 \):
[tex]\[ x^2 (x^3 + 2) = 0 \][/tex]
3. Set Each Factor Equal to Zero: This gives us two separate equations to solve:
[tex]\[ x^2 = 0 \quad \text{and} \quad x^3 + 2 = 0 \][/tex]
4. Solve the First Equation: The first equation \( x^2 = 0 \) has a straightforward solution:
[tex]\[ x = 0 \][/tex]
This solution has a multiplicity of 2, meaning it counts as two solutions.
5. Solve the Second Equation: The second equation \( x^3 + 2 = 0 \) can be solved by isolating \( x \):
[tex]\[ x^3 = -2 \][/tex]
[tex]\[ x = \sqrt[3]{-2} \][/tex]
The cube root of -2 provides three complex roots (one real and two complex). The real root is:
[tex]\[ x = -\sqrt[3]{2} \][/tex]
The remaining two roots are complex.
6. Count All Solutions: Since the polynomial is of degree 5 (the highest exponent is 5), there must be 5 solutions in total. We have:
- Two solutions from \( x = 0 \) (with multiplicity 2)
- Three solutions from \( x^3 = -2 \) (one real and two complex)
Therefore, the polynomial equation \( x^5 = -2x^2 \) indeed has five solutions, considering both real and complex roots.
Conclusion:
The correct statement is:
He is not correct because the greatest exponent of the system is five, so there must be five solutions, three of which must be multiplicities or complex.
1. Rewrite the Equation: Start by rewriting the equation in standard polynomial form:
[tex]\[ x^5 + 2x^2 = 0 \][/tex]
2. Factor Out the Common Term: Factor out the common term \( x^2 \):
[tex]\[ x^2 (x^3 + 2) = 0 \][/tex]
3. Set Each Factor Equal to Zero: This gives us two separate equations to solve:
[tex]\[ x^2 = 0 \quad \text{and} \quad x^3 + 2 = 0 \][/tex]
4. Solve the First Equation: The first equation \( x^2 = 0 \) has a straightforward solution:
[tex]\[ x = 0 \][/tex]
This solution has a multiplicity of 2, meaning it counts as two solutions.
5. Solve the Second Equation: The second equation \( x^3 + 2 = 0 \) can be solved by isolating \( x \):
[tex]\[ x^3 = -2 \][/tex]
[tex]\[ x = \sqrt[3]{-2} \][/tex]
The cube root of -2 provides three complex roots (one real and two complex). The real root is:
[tex]\[ x = -\sqrt[3]{2} \][/tex]
The remaining two roots are complex.
6. Count All Solutions: Since the polynomial is of degree 5 (the highest exponent is 5), there must be 5 solutions in total. We have:
- Two solutions from \( x = 0 \) (with multiplicity 2)
- Three solutions from \( x^3 = -2 \) (one real and two complex)
Therefore, the polynomial equation \( x^5 = -2x^2 \) indeed has five solutions, considering both real and complex roots.
Conclusion:
The correct statement is:
He is not correct because the greatest exponent of the system is five, so there must be five solutions, three of which must be multiplicities or complex.
