The roots of the expression $2 x^{4} + x^{3} - 6 x^{2} + x + 2$ are:

\frac{1}{2}, 1, 1, 2

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1, 1, - 2, - \frac{1}{2}

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1, 1, 2, - \frac{1}{2}

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1, 1, - 2, \frac{1}{2}

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Solution

The correct option is B $1, 1, - 2, - \frac{1}{2}$
The given expression is $2 x^{4} + x^{3} - 6 x^{2} + x + 2$ .
Sum of all the coefficients is $0$ , hence one obvious root is $x = 1$ .
Dividing the entire polynomial by $x - 1$ we get
$2 x^{3} + 3 x^{2} - 3 x - 2 = 0$
Here also, the sum of all the coefficients is $0$ .
Hence, one root is again $x = 1$ .
Hence $2 x^{3} + 3 x^{2} - 3 x - 2 = 0$
$(x - 1) (x + 2) (2 x + 1) = 0$
Hence, the roots of the above polynomial are
$x = 1, 1, - 2, \frac{- 1}{2}$ .

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