Question

The polynomial with $1,\frac{1}{2}\&\frac{-1}{2}$ as zeroes is ______

• A. $x^3-4x^2-x+1$
• B. $4x^3+4x^2-x+1$
• C. $4x^3-4x^2-x+1$
• D. $4x^3-4x^2-x-1$

Answer 1

The correct option is C $4x^3-4x^2-x+1$

Given, Let zeores be $\alpha ;\beta \&\gamma$
$\alpha =1;\beta =\frac{1}{2}\&\gamma =\frac{-1}{2}$

We know the cubic polynmial with zeores is given by
$k\left[x^3\right]-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma \alpha +\beta +\gamma =1+\frac{1}{2}-\frac{1}{2}=1\alpha \beta +\beta \gamma +\gamma \alpha =1\times \frac{1}{2}+\frac{1}{2}\times \frac{-1}{2}+1\times \frac{-1}{2}=\frac{1}{2}-\frac{1}{4}-\frac{1}{2}=\frac{2-1-2}{4}=\frac{-1}{4}\alpha \beta \gamma =1\times \frac{1}{2}\times \frac{-1}{2}=\frac{-1}{4}\therefore k[x^3-1x^2-\frac{1}{4}x-\frac{-1}{4}]k\left[\frac{4x^3-4x^2-x+1}{4}\right]4x^3-4x^2-x+1[k=\frac{1}{4}]$

The cubic polynomial is $4x^3-4x^2-x+1$