HELP‼️⚠️‼️⚠️‼️⚠️‼️⚠️‼️ Which polynomial is prime?
O 3x^3+3x^2-2x-2
O 3x^3-2x^2+3x-4
O 4x^3+2x^2+6x+3
O 4x^3+4x^2-3x-3

There is either No Answer or

B See if the computer or your maker will take B, but the real answer is that there is no answer.

Step-by-step explanation:

Prime polynomials cannot be factored.

x^2 + 4 can't be factored.

x^2 - 4 can be factored and is not prime

x^2 - 4 = (x + 2)(x - 2)

A) is not prime

3x^2(x + 1) - 2(x + 1)       x + 1 is a common factor

(3x^2 - 2)(x + 1)

B) Not sure. I'll look at the rest. It has a common divisor (x - 1) which turns the result into 0. I don't think you've likely taken this yet, but the answer is        (x - 1)(3x^2 + x + 1), so although this one is not prime, it is not done the same way as the other three. It likely is the answer, but I sure would be tempted to report it to your instructor.

C) 3x^2(x + 1) - 2(x + 1)   x + 1 is a common factor. C is not the answer.

D D is done the same way that A and C are done.

4x^2(x + 1) - 3(x + 1) and x + 1 is a common factor.

So the answer might be B. It doesn't look like it factors the same way as the other three

D. 2x4+x3–x+2 is prime

Step-by-step explanation:

A prime polynomial is a polynomial that can't be factored. Hope I helped

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°