Find the zeroes of the polynomial …

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Sia ? 6 years, 3 months ago
7y2 - {tex}\frac { 11 } { 3 } y - \frac { 2 } { 3 }{/tex}
= {tex}\frac 13{/tex}(21y2 - 11y - 2)
= {tex}\frac 13{/tex}(21y2 - 14y + 3y - 2)
= {tex}\frac 13{/tex}[7y(3y - 2) + 1(3y - 2)]
= {tex}\frac 13{/tex}(3y - 2)(7y + 1)
{tex}\Rightarrow y = \frac { 2 } { 3 } , \frac { - 1 } { 7 }{/tex} are zeroes of the polynomial.
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