Question

Which of these is a zero of the polynomials p(y) = 3y^3 - 16 y - 8? * - 8 0 2 - 2

Answer 1

The zero of polynomials is -2. The polynomials is p(y) = $3y^{3} - 16 y - 8$.

According to the question,

The polynomials is p(y) = $3y^{3} - 16 y - 8$. In order to find the zero of polynomials only if we substitute the value and polynomials become zero.

p(-2) = $3(-2)^{3} - 16 (-2) - 8$

= 3(8) + 32 -8

= 0

Hence, the zero of polynomials is -2. The polynomials is p(y) = $3y^{3} - 16 y - 8$.

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