Zeroes of the polynomial $p(x)=x^{2}-3 \sqrt{2} x+4$ are :\

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Professor PC Thomas Classes · Answer

(b) : We have, $p(x)=x^{2}-3 \sqrt{2} x+4$\ $\Rightarrow p(x)=x^{2}-2 \sqrt{2} x-\sqrt{2} x+4$ $$ =x(x-2 \sqrt{2})-\sqrt{2}(x-2 \sqrt{2})=(x-2 \sqrt{2})(x-\sqrt{2}) $$ For zeroes of the polynomial, put $p(x)=0$\ $\Rightarrow \quad(x-2 \sqrt{2})(x-\sqrt{2})=0$\ $	herefore \quad x=2 \sqrt{2}, \sqrt{2}$ are zeroes of $p(x)$.\