If the polynomial x raise to power 4 - 6 x cube + 16 x square - 25 x + 10 is divided by another polynomial x square - 2 x + k does remainder comes out we x + a find k and a
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Given that the remainder is (x + a)
⇒ (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
Given that the remainder is (x + a)
⇒ (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
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