If (x-1) is the HCF of Px2-Qx+R and Qx2-Px+R, then what is the value of R?
If (x-1) is the HCF of Px2-Qx+R and Qx2-Px+R, then what is the value of R? Correct Answer 0
(x-1) is the HCF of Px2-Qx+R and Qx2-Px+R. P(12)-Q(1)+R = 0 R = Q-P …… (i) Q(12)-P(1)+R = 0 R = P-Q …… (ii) Solving (i) and (ii), we get, R = 0.
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Feb 20, 2025