4-2.Quadratic Equations and Inequations
easy

यदि $(x + 1)$ व्यंजक ${x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$

का एक गुणनखण्ड हो, तो $p = $

A

$4$

B

$2$

C

$1$

D

इनमें से कोई नहीं

(IIT-1975)

Solution

यदि $(x + 1)$

${x^4} – (p – 3){x^3} – (3p – 5){x^2} + (2p – 7)x + 6$ का गुणांक है तब $x =  – 1$ रखने पर

$1 + (p – 3) – (3p – 5) – (2p – 7) + 6 = 0$

$\Rightarrow$  $ – 4p =  – 16\,\,\, \Rightarrow p = 4$.

Standard 11
Mathematics

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