Obtain all other zeroes or the polynomial $x^{4} + 6 x^{3} + x^{2} - 24 x - 20,$ if two of its zeroes are + 2 and -5.

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Solution

Given, 2, - 5 are the zeroes of polynomial
Let $p (x) = x^{4} + 6 x^{3} + x^{2} - 24 x - 20$
So (x - 2) and (x + 5) are factors of p(x)
$\Rightarrow (x - 2) (x + 5)$ is also a factor of p(x)
So $(x - 2) (x + 5) = x^{2} + 3 x - 10$
$x^{2} + 3 x - 10) x^{4} + 6 x^{3} + x^{2} - 24 x - 20 (x^{2} + 3 x + 2$
$\begin{matrix} x^{4} + 3 x^{3} - 10 x^{2} - - + - ----------------------- - 3 x^{3} + 11 x^{2} - 24 x - 20 3 x^{3} + 9 x^{2} + - 30 x - - + - ------------------ - 2 x^{2} + 6 x - 20 2 x^{2} + 6 x - 20 - - + - ------------------ - 0 \end{matrix}$
So, by remainder theorem,
Dividend = Divisor $\times$ Quotient + Remainder
$x^{4} + 6 x^{3} + x^{2} - 24 x - 20 = (x^{2} + 3 x - 10) \times (x^{3} + 3 x + 2) + 0$
$= (x^{2} + 3 x - 10) (x^{2} + 2 x + x + 2)$
$= (x^{2} + 3 x - 10) [x (x + 2) + 1 (x + 2)]$
$= (x^{2} + 3 x - 10) (x + 2) (x + 1)$
So other zeros are -2 and -1.

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