A polynomial function \( p \) is given by \( p(x)=-x(x-4)(x+2) \), what are all intervals on which \( p(x) \geq 0 \) ? (A) \( [-2,4] \) (B) \( [-2,0] \cup[4, \infty) \) (C) \( (-\infty,-4] \cup[0,2] \) (D) \( (-\infty,-2] \cup[0,4] \)

To find the intervals on which the polynomial function \( p(x) = -x(x-4)(x+2) \) is greater than or equal to 0, we need to determine the sign of the function in different intervals. Given function: \( p(x) = -x(x-4)(x+2) \) To find the intervals where \( p(x) \geq 0 \), we need to consider the sign of the function in different intervals. The critical points of the function are the values of \( x \) where the function changes sign. These critical points are the roots of the function. The roots of the function are the values of \( x \) that make the function equal to zero. So, we need to find the roots of the function \( p(x) = -x(x-4)(x+2) \). The roots of the function are: 1. \( x = 0 \) 2. \( x = 4 \) 3. \( x = -2 \) Now, we will test the sign of the function in different intervals created by these roots. 1. For \( x < -2 \): - Choose a test point, say \( x = -3 \) - Substitute \( x = -3 \) into the function: \( p(-3) = -(-3)(-3-4)(-3+2) = -(-3)(-7)(-1) = 21 \) - Since \( p(-3) > 0 \), the function is positive in the interval \( x < -2 \). 2. For \( -2 < x < 0 \): - Choose a test point, say \( x = -1 \) - Substitute \( x = -1 \) into the function: \( p(-1) = -(-1)(-1-4)(-1+2) = -(-1)(-5)(1) = 5 \) - Since \( p(-1) > 0 \), the function is positive in the interval \( -2 < x < 0 \). 3. For \( 0 < x < 4 \): - Choose a test point, say \( x = 1 \) - Substitute \( x = 1 \) into the function: \( p(1) = -(1)(1-4)(1+2) = -(1)(-3)(3) = 9 \) - Since \( p(1) > 0 \), the function is positive in the interval \( 0 < x < 4 \). 4. For \( x > 4 \): - Choose a test point, say \( x = 5 \) - Substitute \( x = 5 \) into the function: \( p(5) = -(5)(5-4)(5+2) = -(5)(1)(7) = -35 \) - Since \( p(5) < 0 \), the function is negative in the interval \( x > 4 \). Therefore, the intervals on which \( p(x) \geq 0 \) are: - \( x < -2 \) - \( -2 < x < 0 \) - \( 0 < x < 4 \) So, the correct answer is (D) \( (-\infty,-2] \cup[0,4] \).