If $x^{2} - 3 x + 2$ be factor of $x^{4} - p x^{2} + q$ , then $(p, q)$ is equal to :

(3, 4)

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(4, 5)

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(4, 3)

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(5, 4)

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Solution

The correct option is A $(3, 4)$
A] Factors of $x^{2} - 3 x + 2$ are (x-1)(x-2)
$∴$ 1,2 are zeros of $x^{4} - p x^{2} + q$
x=1,
$(1) - p (1) + q = 0 \Rightarrow 1 + q = p - - - (1)$
x=2
$(2^{4}) - p (2)^{2} + q = 0 \Rightarrow 16 - 4 p + q = 0 - - - (2)$
solving (1) & (2)
16-4(1+q)+q=0
=16-4-4q+q=0
=12-3q = 0
$∴$ q=3
p = q + 1 = 4
(p,q) = (4,3)

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If x2−3x+2 be factor of x4−px2+q, then (p,q) is equal to :

The correct option is A (3,4)A] Factors of x2−3x+2 are (x-1)(x-2)∴ 1,2 are zeros of x4−px2+qx=1,(1)−p(1)+q=0⇒1+q=p−−−(1)x=2(24)−p(2)2+q=0⇒16−4p+q=0−−−(2)solving (1) & (2)16-4(1+q)+q=0=16-4-4q+q=0=12-3q = 0∴ q=3p = q + 1 = 4(p,q) = (4,3)

If $x^{2} - 3 x + 2$ be factor of $x^{4} - p x^{2} + q$ , then $(p, q)$ is equal to :