Question

Find the values if a and b so that x^4 +x^3 +8x^2 +ax -b is divisible by x^2 +1

Answer 1

                 

$p(x) = x^4 +x^3 +8x^2 +ax -b = 0$

$If \ p(x)\ is divisible by \ x^2+1,\ then \ p(\iota)=0\ and \ p(-\iota)=0,\ where \ \iota=\sqrt{-1}\ .$

$p(\iota)=(\iota)^4+(\iota)^3+8(\iota)^2+a(\iota)-b=0 \\ \\ p(\iota)=1-\iota-8+a\iota-b=0\ \ \ \ \ \longrightarrow\ \ \ \ \ (1)$

$p(-\iota)=(-\iota)^4+(-\iota)^3+8(-\iota)^2+a(-\iota)-b=0 \\ \\ p(-\iota)=1+\iota-8-a\iota-b=0\ \ \ \ \ \longrightarrow\ \ \ \ \ (2)$

$(1) = (2) \\ \\ 1-\iota-8+a\iota-b=1+\iota-8-a\iota-b \\ \\ -\iota+a\iota=\iota-a\iota \\ \\ \iota(a-1)=\iota(1-a) \\ \\ a-1=1-a \\ \\ a+a=1+1 \\ \\ 2a=2 \\ \\ a=\bold{1}$

$From \ (1), \\ \\ 1-\iota-8+a\iota-b=0 \\ \\ 1-\iota-8+\iota-b=0 \\ \\ -7-b=0 \\ \\ b=\bold{-7}$

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