Question

find a cubic polynomial whose zeros are 1/2, -1/2, 1 ​

Answer 1

$\huge \boxed{ \bf{Answer}}$


Let the zeroes of the cubic polynomial be $\bf{\alpha}$ , $\bf{\beta}$ and $\bf{\gamma}$ such that,


$\alpha = \frac{1}{2} \: \: \: \beta = \frac{ - 1}{2} \: \: and \: \: \gamma = 1$


Now ,


$= > \: \alpha + \beta + \gamma = \frac{ - b}{a}$

$= > \: \frac{1}{2} + \frac{ - 1}{2} \: + 1 = \frac{ - b}{a}$

$= > \frac{ - b}{a} = 1$

$= > - 1 = \frac{b}{a}$


$= > \: \alpha \beta + \beta \gamma + \gamma \alpha \: = \frac{c}{a}$

$= > \: (\frac{1}{2})( \frac{ - 1}{2}) + (\frac{ - 1}{2} )(1) + (1)( \frac{1}{2} ) = \frac{c}{a}$

$= > \frac{ - 1}{4} \: - \: \frac{1}{2} + \frac{1}{2} \: = \frac{c}{a}$

$= > - \frac{1}{4} \: = \frac{c}{a}$


$= > \: \alpha \times \beta \times \gamma \: = \: \frac{ - d}{a}$

$= > \: \frac{1}{2} \times \frac{ - 1}{2} \times 1 = \frac{ - d}{a}$

$= > \: - \frac{1}{4} = \frac{ - d}{a}$

$= > \: \frac{1}{4} = \frac{d}{a}$


So, by placing the values of the zeroes respectively, we get the cubic polynomial as

${x}^{3} - {x}^{2} - \frac{1}{4}x \: + \: \frac{1}{4}$


$\therefore \boxed{ \bf{\: Required \: Answer \: = {x}^{3} - {x}^{2} - \frac{1}{4} x + \frac{1}{4} }}$