Find the number of polynomials of the form x^3+ax^2+bx+c that are divisible by x^2+1 ?
where a,b,c belong to (1,2,3,4---------10).

• A: x^3+ax^2+bx+c = (x^3+bx)+(ax^2+c).. now b has be 1 to get (x^2+1) as a common factor.. now a=1,c=1 is one solution.. hence its, (x^3+x)+(x^2+1)=x(x^2+1)+(x^2+1)=(x^2+1)(x+1) which is divisible by x^2+1..
  values possible:
  b=1, a=1, c=1
  b=1, a=2, c=2
  b=1, a=3, c=3
  b=1, a=4, c=4
  b=1, a=5, c=5
  b=1, a=6, c=6
  b=1, a=7, c=7
  b=1, a=8, c=8
  b=1, a=9, c=9
  b=1, a=10, c=10
  So.. Total of 10 solutions..
• 10 years agoHelpfull: Yes(73) No(5)
• 10 solutions.
  bcz when we divide by x^2+1 we get (b-1)x+(c-a) as remainder, and for x^2+1 to divide the given polynomial remainder should be zero. So to make remainder 0, b=1 and a=c, therefore 10 solutions possible.
• 10 years agoHelpfull: Yes(39) No(11)
• 10 solutions
• 10 years agoHelpfull: Yes(1) No(19)
• on dividing the polynomial we get the remainder (b-1)x+(c-a) as it should be completely divisible so,
  b-1=0 => b=1
  and c-a=0 => c=a
  so total 10 values will satisfy them as the range is given from 1,2,3...10.
• 7 years agoHelpfull: Yes(0) No(0)