Given that x = 2 and x = -1 are zeros of a polynomial function of degree 3, with multiplicities 1 and 2 respectively, the polynomial function can be written in the form of f(x) = a(x - 2)(x + 1)^2, where a is a constant.

We are also given that f(0) = 6. We can substitute x = 0 into the function to solve for a:

6 = a(0 - 2)(0 + 1)^2
6 = a(-2)(1)
6 = -2a
a = -3

Therefore, the polynomial function is f(x) = -3(x - 2)(x + 1)^2.

So, the correct answer is D. -3(x + 1)^2(x - 2).

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Given that x = 2 and x = -1 are zeros of a polynomial function of degree 3, with multiplicities 1 and 2 respectively, the polynomial function can be written in the form of f(x) = a(x - 2)(x + 1)^2, where a is a constant.

We are also given that f(0) = 6. We can substitute x = 0 into the function to solve for a:

6 = a(0 - 2)(0 + 1)^2
6 = a(-2)(1)
6 = -2a
a = -3

Therefore, the polynomial function is f(x) = -3(x - 2)(x + 1)^2.

So, the correct answer is D. -3(x + 1)^2(x - 2).

Knowee AI · Accepted Answer

Given that x = 2 and x = -1 are zeros of a polynomial function of degree 3, with multiplicities 1 and 2 respectively, the polynomial function can be written in the form of f(x) = a(x - 2)(x + 1)^2, where a is a constant.

We are also given that f(0) = 6. We can substitute x = 0 into the function to solve for a:

6 = a(0 - 2)(0 + 1)^2
6 = a(-2)(1)
6 = -2a
a = -3

Therefore, the polynomial function is f(x) = -3(x - 2)(x + 1)^2.

So, the correct answer is D. -3(x + 1)^2(x - 2).

Given x=2 and x=-1 anre zeros with multiplicity 1 and two respectively of a polynomimal function of degree 3. If f(0) = 6, then f(x)=A.(x+1)2(x-2)B.-3(x+1)2(x+2)C.3(x+1)2(x-2)D.-3(x+1)2(x-2)

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