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EXPLANATION:

\huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Question:-}}}}}}}

\bullet{\longmapsto} Whether \sf \dfrac{1}{2} and 1 are zeros of the POLYNOMIAL p(X) = 2x² - 3x + 1 or not? JUSTIFY.

\huge {\bf{\orange{\mathfrak{\dag{\underline{\underline{Answer:-}}}}}}}

\bullet{\leadsto} \: \sf p(x) \: = \: 2x^{2} \: - \: 3x \: + \: 1.

\bullet{\leadsto} \: \sf Check \: whether \: \dfrac{1}{2} \: and \: 1 \: are \: zeros?

\bullet{\leadsto} \: \sf So, \: we \: have \: to \: substitute \: given \: values \: in \: p(x).

\bullet{\leadsto} \: \sf p(x) \: = \: 2x^{2} \: - \: 3x \: + \: 1.

\bullet{\leadsto} \: \sf Substituting \: x \: = \: \dfrac{1}{2} \: in \: p(x).

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: 2 \: * \: (\dfrac{1}{2})^{2} \: - \: 3 \: * \: \dfrac{1}{2} \: + \: 1.

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: 2 \: * \: \dfrac{1}{4} \: - \: 3 \: * \: \dfrac{1}{2} \: + \: 1.

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: \cancel{2} \: * \: \dfrac{1}{\cancel{4}} \: - \: 3 \: * \: \dfrac{1}{2} \: + \: 1.

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: \dfrac{1}{2} \: - \: 3 \: * \: \dfrac{1}{2} \: + \: 1.

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: \dfrac{1}{2} \: - \: \dfrac{3}{2} \: + \: 1.

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: 0.5 \: - \: 1.5 \: + \: 1.

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: 1 \: + \: 1.

\bullet{\leadsto} \: \sf p(\dfrac{1}{2}) \: = \: 2.

\bullet{\leadsto} \: \underline{\boxed{\pink{\sf p(\dfrac{1}{2}) \: = \: 2 \: \neq \: 0.}}}

\bullet{\leadsto} \: \sf Substituting \: x \: = \: 1 \: in \: p(x).

\bullet{\leadsto} \: \sf p(1) \: = \: 2 \: * \: 1^{2} \: - \: 3 \: * \: 1 \: + \: 1.

\bullet{\leadsto} \: \sf p(1) \: = \: 2 \: * \: 1 \: - \: 3 \: + \: 1.

\bullet{\leadsto} \: \sf p(1) \: = \: 2 \: - \: 3 \: + \: 1.

\bullet{\leadsto} \: \sf p(1) \: = \: - \: 1 \: + \: 1.

\bullet{\leadsto} \: \underline{\boxed{\green{\sf p(1) \: = \: 0.}}}

\huge{\bf{\red{\mathfrak{\dag{\underline{\underline{Conclusion:-}}}}}}}

\bullet{\longmapsto} \: \boxed{\therefore{\sf Zeros \: of \: polynomial \: p(x) \: = \: 2x^{2} \: - \: 3x \: + \: 1 \: is \: 1 .}}


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