Question

What is the zero's of the polynomial p(y)=3y^2-1

Answer 1

$Given \: polynomial \: p(y) = 3y^{2} - 1$

/* To find zeroes p(y) , we take p(y) = 0 */

$3y^{2} - 1 = 0$

$\implies 3y^{2} = 1$

$\implies y^{2} = \frac{1}{3}$

$\implies y = \pm \Big( \frac{1}{3}\Big)$

Therefore.,

$\pm { \frac{1}{\sqrt{3}} \:are \: zeroes \: p(y)}$

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