x^4-7x^3+21x^2+ax+b = 0 has a root 1+2i
.
Therefore the conjugate of 1+2i is also a root of the
given equation.
The conjugate of 1+2i is
1-2i.
Therefore , (x-(1+2i)) and (x-(1-2i)) are the factors
of x^4-7x^3+21x^2+ax+b by remainder
theorem.
(x-1-2i)(x-1+2i) = (x-1)^2 - (2i)^2 = (x^2-2x+1+4)
= (x^2+5x+5) is also a factor.
Therefore the x^2-7x^3+21x^2
+ax+b should be divisible by (x^2-2x+5).
So we divide
x^4-7x^3+21x^2+ax+b by
x^2-2x+5.
x^2-2x+5)x^4-7x^3+21x+ax+b(
x^2
x^4 -2x^3
+5x^2
------------------------------------------
x^2-2x+5)
-5x^3+16x^2 +ax+b ( x^2-5x
-5x^3 +10x^2
-25x
-------------------------------------------------
x^2-2x+5)6x^2+(a+25)x+b
(x^2-5x+6
6x^2 -12x
+30
---------------------------------------
(
a+25 +12)x +(b-30)
So the remainder = (a+37)x +(b-30)
.
Since x^2-2x+5 is factor of x^4-7x^3+21x^2+ax+b, the
remainder (a+37)x +(b-30) should be zero.
Therefore the
coefficient of x , a+37= 0 and the contant term b-30 =
0.
Therefore a = -37 and b =
30.
So the given equation should be
x^2-7x^3+21x^2-37x+30
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