If the polynomial x ki power …
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Gaurav Seth 3 years, 9 months ago
We know that,
Dividend = Divisor × Quotient + Remainder
⇒ Dividend – Remainder = Divisor × Quotient
⇒ Dividend – Remainder is always divisible by the divisor.
Now, it is given that f(x) when divided by x2 – 2x + k leaves (x + a) as remainder.
So, for f(x) to be completely divisible by x2 – 2x + k, remainder must be equal to zero
⇒ (–10 + 2k)x + (10 – a – 8k + k2) = 0
⇒ –10 + 2k = 0 and 10 – a – 8k + k2 = 0
⇒ k = 5 and 10 – a – 8 (5) + 52 = 0
⇒ k = 5 and – a – 5 = 0
⇒ k = 5 and a = –5
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