If the degree of ((c+2)x^5+(c^2-4)x^3+9) is (0), what is the value of (c)?

Correct answer: (c=-2)

The direct answer is option B, \(c=-2\). A polynomial has degree 0 when it is a nonzero constant, so every term containing x must disappear. The coefficient of \(x^5\) is \(c+2\). Setting it to zero gives \(c+2=0\), hence \(c=-2\). Check the other coefficient: \(c^2-4=(-2)^2-4=4-4=0\). The expression therefore becomes only \(9\), whose degree is 0. Option B is correct. Option A, \(c=2\), makes the first coefficient 4, so an \(x^5\) term remains. Option C, \(c=0\), gives coefficients 2 and -4, so the degree is still 5. Option D, \(c=4\), also leaves both variable terms present. The safe method is to make the coefficients of all positive powers zero and then check that the constant term is not zero.