The polynomial ky3+3y2-3 and 2y3-5y+k when divided by y-5 leave the same remainder in each case find the value of k
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y-5 = 0
y = 5
p(5)= ky³+3y²-3 = 2y³-5y+k
k(5)³+3(5)²-3 = 2(5)³-5(5)+k
125k+75-3=250-25+k
125k-k=225-72
124k = 153
k = 153/124
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Answer:153/124
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