CDS 2026 (I) Elementary Mathematics, question 81

Topic: Algebra

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Passage · Q81 to Q82

$x^{4}+px^{3}+qx^{2}+x+6$ is divisible by $x^{2}-x-6$ .

Q81

What is the value of $p$ ?

Solution.

Answer: (b) $-1$

The divisor $x^{2} - x - 6$ factors as $(x - 3)(x + 2)$ , so the polynomial F must vanish at 3 and $- 2$ .
$F(3) = 0$ gives $90 + 27p + 9q = 0$ , or $3p + q = - 10$ .
$F( - 2) = 0$ gives $20 - 8p + 4q = 0$ , or $- 2p + q = - 5$ .
Subtracting yields $5p = - 5$ .