X^4+x^3+8x^2+ax+b is divisible by x^2+1. Find the value of \'a\' &\'b\

1.	X^4+x^3+8x^2+ax+b is divisible by x^2+1. Find the value of \'a\' &\'b\'
Answer» If {tex}x^4+x^3+8x^2+ax +b{/tex}\xa0is exactly divisible by x2 + 1, the remainder after division should be zero.Now let us perform long divisionWe get, remainder = x (a - 1) + (b - 7)\xa0x (a - 1) + (b - 7 ) = 0{tex}\\Rightarrow{/tex}\xa0x (a - 1) + (b - 7) = 0x + 0{tex}\\Rightarrow{/tex}\xa0a - 1 = 0 and b - 7 = 0\xa0[On equating the coefficients of like powers of x]{tex}\\Rightarrow{/tex}a = 1 and b = 7

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