Find the value of a and b so that x4+x3+8x2+ax +b is exactly divisible by x2+1
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Here's another approach:
Note that x^2 + 1 = (x + i)(x - i).
Since these factors are relatively prime, we need x^4 + x^3 + 8x^2 + ax + b to be divisible by both factors (x + i) and (x - i). (Since a and b are tacitly assumed to be real, we only need to work with one of the two factors; I'll write out details for both anyway.)
By the Factor Theorem, we have
(-i)^4 + (-i)^3 + 8(-i)^2 + a(-i) + b = 0 and
i^4 + i^3 + 8i^2 + ai + b = 0
Simplifying:
1 + i - 8 - ai + b = 0 ==> -ai + b = 7 - i
1 - i - 8 + ai + b = 0 ==> ai + b = 7 + i.
Since a and b are real, we can simply equate real and imaginary parts from either equation and obtain a = 1 and b = 7.
Note that x^2 + 1 = (x + i)(x - i).
Since these factors are relatively prime, we need x^4 + x^3 + 8x^2 + ax + b to be divisible by both factors (x + i) and (x - i). (Since a and b are tacitly assumed to be real, we only need to work with one of the two factors; I'll write out details for both anyway.)
By the Factor Theorem, we have
(-i)^4 + (-i)^3 + 8(-i)^2 + a(-i) + b = 0 and
i^4 + i^3 + 8i^2 + ai + b = 0
Simplifying:
1 + i - 8 - ai + b = 0 ==> -ai + b = 7 - i
1 - i - 8 + ai + b = 0 ==> ai + b = 7 + i.
Since a and b are real, we can simply equate real and imaginary parts from either equation and obtain a = 1 and b = 7.
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