Question

finds remainder ,when x³+3x²-1 is divided by x​

Answer 1

Answer:

x+1=0

x+1=0x=−1

x+1=0x=−1p(x)=x3+3x2−3x−1p(−1)

x+1=0x=−1p(x)=x3+3x2−3x−1p(−1)=(−1)3+3(−1)2−3(−1)−1

x+1=0x=−1p(x)=x3+3x2−3x−1p(−1)=(−1)3+3(−1)2−3(−1)−1=−1+3(1)+3−1

x+1=0x=−1p(x)=x3+3x2−3x−1p(−1)=(−1)3+3(−1)2−3(−1)−1=−1+3(1)+3−1=−1+3+3−1

x+1=0x=−1p(x)=x3+3x2−3x−1p(−1)=(−1)3+3(−1)2−3(−1)−1=−1+3(1)+3−1=−1+3+3−1=6−2

x+1=0x=−1p(x)=x3+3x2−3x−1p(−1)=(−1)3+3(−1)2−3(−1)−1=−1+3(1)+3−1=−1+3+3−1=6−2=4

x+1=0x=−1p(x)=x3+3x2−3x−1p(−1)=(−1)3+3(−1)2−3(−1)−1=−1+3(1)+3−1=−1+3+3−1=6−2=4Thus remainder is 4

Answer 2

Answer:

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