When the polynomial a3+2a2−5ax−7 is divided by a+1, the remainder is R1. If R1=14, find the value of x.
A
1
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B
2
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C
3
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D
4
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Solution
The correct option is D4 When the polynomial a3+2a2−5ax−7 is divided by a+1, the remainder is R1. By the remainder theorem, the remainder is, R1=(−1)3+2(−1)2−5(−1)x−7 14=−1+2+5x−7=5x−6 14=5x−6 14+6=5x 20=5x ∴x=4