Factor Theorem-10

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If (x + a) be the HCF of x2 + mx + n and x2 + rx + s, then show that a = (n - s) / (m - r)

Solution: Let P(x) = x2 + mx + n and Q(x) = x2 + rx + s,

(x + a) is the HCF of both P(x) and Q(x) means, (x + a) is a factor of both P(x) and Q(x). Therefore,

P( -a) = 0; => ( -a)2 + m( -a) + n = 0; => a2 -am + n = 0 -------------------------------(i)

Q( -a) = 0; => ( -a)2 + r( -a) + s = 0; => a2 -ar + s = 0 -------------------------------(ii)

By (i) – (ii), we get

ar – am + n – s = 0

=> am – ar = n – s

=> a(m - r) = n – s

=> a = (n - s) / (m - r)

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