Question

find the value of a and b if x^4+x^3+6x^2+ax+b is divisible by x^2+1

Answer 1

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Answer 2

{ b -> 5, a -> 1}

Answer 3

if a is replaced by 1 and b is replaced by 4 then the expression is:\[x^4+x^3+6 x^2+x+5 \]If the above is divided by x^2+1 then the remainder is zero.
From Mathematica:\[\text{PolynomialQuotientRemainder}\left[x^4+x^3+6 x^2+x+5,x{}^{\wedge}2+1,x\right]= \]\[\left\{x^2+x+5,0\right\} \]

Answer 4

The quotient and remainder from dividing the problem expression by x^2+1 is:\[\left\{5+x+x^2,-5+b+(-1+a) x\right\} \]The quotient has to be equal to zero. if b is 5 and
a is one the remainder evaluates to zero.

Answer 5

Sorry. "The quotient has to be equal to zero" should have been typed as "The remainder has to be equal to zero"