Question

Find all the other zeroes of the polynomial p(x)= x^4+3x^3-x^2-9x-6 if two of zeroes are √3 and -√3

Answer 1

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Answer 2

$p(x) = {x}^{4} + 3 {x}^{3} - {x}^{2} - 9x - 6$
Given that two the zeroes of p(x) are
$\sqrt{3} \text {and} -\sqrt{3}$.

From Factor Theorem,

$(x - \sqrt{ 3} )$
and
$(x + \sqrt{3} )$
are factors of p(x).

Or
$(x - \sqrt{3} )(x + \sqrt{3} ) = {x}^{2} - 3$
is also a factor of p(x).

Now, by Dividing:

$( {x}^{2} - 3)(x +1)(x + 2) = {x}^{4} - 3 {x}^{3} - {x}^{2} - 9x - 6$
Now,
$p(x) = 0$
Thus,

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

The first two roots are already given,
Thus the other two roots are
$-1$
and
$-2$