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Find all the zeroes of the polynomial $p (x) = 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2$ if you know the two zeroes are $\sqrt{2}$ and $\sqrt{- 2}$ .

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

\sqrt{3}, - \sqrt{5}, 2, and \frac{3}{2}

\sqrt{- 2}, \sqrt{3}, 3, and \frac{5}{2}

\sqrt{5}, \sqrt{7}, - 5, and \frac{1}{3}

answer is A.

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Detailed Solution

Given polynomial equation is,

2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2

Its zeros are

\sqrt{2}

and

\sqrt{- 2}

So, two factors are

(x - \sqrt{2}) and (x + \sqrt{2})

So, the product of the factors,

(x - \sqrt{2}) \times (x + \sqrt{2}) = x^{2} - 2

Thus, using long division method divide given polynomial by the above expression as

,

So, the result we get is

2 x^{2} - 3 x + 1

Its factor,

2 x^{2} - 3 x + 1

= 2 x^{2} - 2 x - x + 1

= 2 x (x - 1) - 1 (x - 1)

= (2 x - 1) (x - 1)

Multiply all the zeros and equate to the zero as,

(x - \sqrt{2}) \times (x + \sqrt{2}) \times (x - 1) \times (2 x - 1) = 0

\Rightarrow x - \sqrt{2} = 0, x + \sqrt{2} = 0, x - 1 = 0, 2 x - 1 = 0 \Rightarrow x = \sqrt{2}, x = - \sqrt{2}, x = 1, 2 x = 1 \Rightarrow x = \sqrt{2}, x = - \sqrt{2}, x = 1, x = \frac{1}{2}

Hence, the four zeroes of

2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2

are

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

Option 1 is correct.

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