If $x^{2} - 3 x + 2$ be a factor of $x^{4} - p x^{2} + q$ , then $(p, q) =$

(5, 4)

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(4, 3)

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(4, 5)

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(3, 4)

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Solution

The correct option is A $(5, 4)$
$x^{2} - 3 x + 2$ be factor of $x^{4} - p x^{2} + q = 0$

Hence, $(x^{2} - 3 x + 2) = 0$ $\Rightarrow (x - 2) (x - 1) = 0$

$\Rightarrow x = 2, 1$ putting these values in given equation

So, $4 p - q - 16 = 0 . . . (i)$
And $p - q - 1 = 0 . . . (i i)$

Solving $(i)$ and $(i i)$ , we get $(p, q) = (5, 4)$

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If x2−3x+2 be a factor of x4−px2+q, then (p,q)=

The correct option is A (5,4)x2−3x+2 be factor of x4−px2+q=0 Hence, (x2−3x+2)=0 ⇒(x−2)(x−1)=0 ⇒x=2,1 putting these values in given equation So, 4p−q−16=0 ...(i) And p−q−1=0 ...(ii) Solving (i) and (ii), we get (p,q)=(5,4)

If $x^{2} - 3 x + 2$ be a factor of $x^{4} - p x^{2} + q$ , then $(p, q) =$