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Question

Expand 1x45x3+7x2+x8 in descending powers of x to four terms, and find remainder.

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Solution

Using long division method for expanding we get,
x4+5x5+18x6+54x7–––––––––––––––––––––––––––––
x45x3+7x2+x8)1
15x1+7x2+x38x4
()(+)()()(+)–––––––––––––––––––––––––––––––––
5x17x2x3+8x4
5x125x2+35x3+5x440x5
()(+)()()(+)––––––––––––––––––––––––––––––––––––––
18x236x3+3x4+40x5
18x290x3+126x4+18x5144x6
()(+)()()(+)––––––––––––––––––––––––––––––––––––––––
54x3123x4+22x5+144x6
54x3270x4+378x5+54x6432x7
()(+)()()(+)–––––––––––––––––––––––––––––––––––––––––––
147x4356x590x6432x7

Hence, remainder = 147x4356x5+90x6+432x7

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