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If the product of two zeros of the polynomial f(x) = 2x3 + 6x2 – 4x + 9 is 3, then its third zero is
  • a)
    9/2
  • b)
    -3/2
  • c)
    3/2
  • d)
    -9/2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If the product of two zeros of the polynomial f(x) = 2x3+ 6x2– 4...
f(x)=2x^3+6x^2–4x+9

a=2, b=6, c=-4, d=9

Let the zeroes of f(x) be x, y, z

Product of zeroes = -d/a

x*y*z = -9/2

3z = -9/2 (given x*y=3)

z = -3/2

Third zero = -3/2
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Most Upvoted Answer
If the product of two zeros of the polynomial f(x) = 2x3+ 6x2– 4...
+ 4x is 2, find the third zero.

We can start by using Vieta's formulas, which state that the sum of the zeros of a polynomial is equal to -b/a and the product of the zeros is equal to c/a, where a, b, and c are the coefficients of the polynomial.

In this case, a = 2, b = 6, and c = 4, so we have:

Sum of zeros = -b/a = -6/2 = -3

Product of zeros = c/a = 4/2 = 2

Let the three zeros be r, s, and t. We know that rs = 2, so we can write:

t = 2/rs

t = 2/(rs)

t = 2/(2)

t = 1

Therefore, the third zero is 1.
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