If the product of two zeroes of the polynomial f(x) = 2x3 + 6x2 – 4x – 9 is 3, then its third zero isSelect an answerA −32 B +32 C 92 D −92

The question is asking for the third zero of the polynomial f(x) = 2x^3 + 6x^2 - 4x - 9, given that the product of the other two zeroes is 3.

The sum and product of the roots of a cubic equation ax^3 + bx^2 + cx + d = 0 are given by the formulas:

Sum of roots = -b/a Product of roots = -d/a

In this case, a = 2, b = 6, c = -4, and d = -9.

Let's denote the three roots of the polynomial as r1, r2, and r3. We know that r1*r2 = 3, and we're looking for r3.

The sum of the roots is -b/a = -6/2 = -3. This means that r1 + r2 + r3 = -3.

Since we know that r1*r2 = 3, we can say that r1 and r2 are both positive (since the product of two negative numbers is also positive). Therefore, r3 must be negative.

We can find r3 by subtracting the sum of r1 and r2 from the total sum of the roots. Since r1 + r2 = -3 - r3 and r1*r2 = 3, we can solve for r3:

r3 = -3 - (r1 + r2) = -3 - sqrt(r1*r2) = -3 - sqrt(3) = -3.732

Therefore, the third root of the polynomial is approximately -3.732. This is not an option in the given answers, so there may be a mistake in the question or the provided answer choices.