Question

a polynomial function has a root of 0 with multiplicity 1, and a root of 2 with multiplicity 4. if the function has a negative leading coefficient, and is of odd degree, which of the following are true? choose three correct answers. the function is negative on (0, 2) the function is positive on (0, ∞). the function is positive on (-∞, 0). the function is negative on (2, ∞).

Explanation:

Step 1: Determine the polynomial form

The polynomial can be written as \(f(x)= -a x(x - 2)^4\) where \(a>0\) (since the leading coefficient is negative and we can factor out the roots with their multiplicities. The degree of the polynomial is \(1 + 4=5\), which is odd).

Step 2: Analyze the sign of the polynomial in different intervals

• Interval \((-\infty, 0)\):

Let \(x=-1\) (a test point in \((-\infty, 0)\)). Then \(f(-1)=-a\times(-1)\times(-1 - 2)^4=-a\times(-1)\times(-3)^4=-a\times(-1)\times81 = 81a\). Since \(a > 0\), \(f(-1)>0\). So the function is positive on \((-\infty, 0)\).

• Interval \((0, 2)\):

Let \(x = 1\) (a test point in \((0, 2)\)). Then \(f(1)=-a\times1\times(1 - 2)^4=-a\times1\times(-1)^4=-a\times1\times1=-a\). Since \(a>0\), \(f(1)<0\). So the function is negative on \((0, 2)\).

• Interval \((2, \infty)\):

Let \(x = 3\) (a test point in \((2, \infty)\)). Then \(f(3)=-a\times3\times(3 - 2)^4=-a\times3\times1^4=-3a\). Since \(a > 0\), \(f(3)<0\). So the function is negative on \((2, \infty)\).

• For the interval \((0,\infty)\), we can see from the above analysis that on \((0, 2)\) the function is negative and on \((2,\infty)\) the function is negative, so the function is not positive on \((0,\infty)\).

Answer:

1. The function is negative on \((0, 2)\)
2. The function is positive on \((-\infty, 0)\)
3. The function is negative on \((2, \infty)\)