Question

Find all the zeros of the polynomial 2x3 + x2 – 6x – 3, if two of its zeros are – √ 3 and √3.

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Answer 1

Answer:

There are 3 zeros

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Answer 2

Answer:

$p(x) = 2x ^{3} + x {}^{2} - 6x - 3$

$\alpha = - \sqrt{3}$

$\beta = \sqrt{3}$

$g(x) = x - \sqrt{3}$

$g(x) = x + \sqrt{3}$

$(x - \sqrt{3)} \: \: \times (x + \sqrt{3)}$

$(x {}^{2} - \sqrt{3 {}^{2} } )$

$x {}^{2} - 3 \: is \: \: also \: \: a \: \: factor$

Now for another zero we divide the polynomial by the above factor,

We get,

$p(x) = 2x {}^{3} + x {}^{2} - 6x - 3$

$g(x) = x {}^{2} - 3$

$q(x) = 2x + 1$

$r(x) = 0$

Now,

$g(x) \times q(x)$

$(x {}^{3} - 3) \times(2x + 1) = 0$

We get,

$x = \frac{ - 1}{2}$

Therefore,

The zeroes of the polynomial are:

$- \sqrt{3}$

$\sqrt{3}$

$\frac{ - 1}{2}$

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