Question

whether 1/2 and 1 are zeros of the polynomial p(X)=2x²-3x+1 or not? Justify.​

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

$p(x) = {2x}^{2} - 3x + 1 \\ p( \frac{1}{2} ) = 2( \frac{1}{2} {)}^{2} - 3( \frac{1}{2} ) + 1$

$\\ = 2 \times (\frac{1}{2} {)}^{2} - \frac{3}{2} + 1 \\ = \frac{1}{2} - \frac{3 + 1}{2} \\ = \frac{ - 2}{2} + 1 \\ = - 1 + 1 \\ = 0 \\ \frac{1}{2} is \: \: a \: \: zero \: \: of \: \: p(x)$