41. $f(x)=4x^{3}-12x^{2}-16x$ a. the numbers 1, -4, and 0 are zeros of multiplicity 2 b. the numbers -1, 4…

41. $f(x)=4x^{3}-12x^{2}-16x$ a. the numbers 1, -4, and 0 are zeros of multiplicity 2 b. the numbers -1, 4, and 0 are zeros of multiplicity 2 c. the numbers -1, 4, and 0 are zeros of multiplicity 1 d. the numbers 1, -4, and 0 are zeros of multiplicity 1

41. $f(x)=4x^{3}-12x^{2}-16x$ a. the numbers 1, -4, and 0 are zeros of multiplicity 2 b. the numbers -1, 4, and 0 are zeros of multiplicity 2 c. the numbers -1, 4, and 0 are zeros of multiplicity 1 d. the numbers 1, -4, and 0 are zeros of multiplicity 1

Answer

Explanation:

Step1: Factor the polynomial

First, factor out the greatest common factor (4x) from (f(x)=4x^{3}-12x^{2}-16x). So, (f(x) = 4x(x^{2}-3x - 4)). Then, factor the quadratic (x^{2}-3x - 4) using the formula (x^{2}+(a + b)x+ab=(x + a)(x + b)). Here, (a=-4) and (b = 1), so (x^{2}-3x - 4=(x-4)(x + 1)). So, (f(x)=4x(x - 4)(x+1))

Step2: Determine the zeros and their multiplicities

Set (f(x)=0), then (4x(x - 4)(x + 1)=0). Using the zero - product property (ab=0\Rightarrow a = 0) or (b=0). If (4x=0), then (x = 0); if (x-4=0), then (x = 4); if (x + 1=0), then (x=-1). Since the polynomial is factored as a product of linear factors (4x(x - 4)(x + 1)), each factor (x), ((x - 4)) and ((x + 1)) has an exponent of (1).

Answer:

B. the numbers (-1), (4), and (0) are zeros of multiplicity (1)