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Nature of Roots

If x+1 is a...

Question

If $x + 1$ is a factor of $x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6$ then find the value of $p$ .

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Solution

Given, $x + 1$ is a factor of $x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6$ .
Then $x = - 1$ will satisfy $x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6 = 0.$
Then we'll get,
$1 + (3 - p) - (3 p - 5) + (9 - 2 p) + 6 = 0$
or, $24 - 6 p = 0$
or, $p = 4$ .

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