Find the value of $p$ for which $x + 1$ is a factor of $x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6$ . Also find the remaining factors for this value of $p$ .

4; x - 2

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4; x - 1

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6; x - 3

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6; x + 3

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Solution

The correct options are
A $4; x - 2$
B $4; x - 1$
Let $f (x) = x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6.$
Given $x + 1$ is a factor of $f (x)$ , $\Rightarrow f (- 1) = 0$
$∴ p = 4$
Thus $x^{4} + x^{3} - 7 x^{2} - x + 6 = (x + 1) (x^{3} - 7 x + 6)$
$= (x + 1) (x - 1) (x^{2} + x - 6)$
$= (x + 1) (x - 1) (x + 3) (x - 2)$
The other factors are $x - 1, x - 2, x + 3.$

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