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Let First polynomial be p(X)
and second be f(X)
Now, f(X) × Q + (x+a) = p(X)
and second be f(X)
Now, f(X) × Q + (x+a) = p(X)
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⇒ (4k – 25 + 16 – 2k)x + [10 – k(8 – k) ] = x + a
⇒ (2k – 9)x + [10 – 8k + k2 ] = x + a
On comparing both the sides, we get
2k – 9 = 1
⇒ 2k = 10
∴ k = 5
Also 10 – 8k + k2 = a
⇒ 10 – 8(5) + 52 = a
⇒ 10 – 40 + 25 = a
∴ a = – 5