Asked by Crow
a and b are integers such that the polynomial f(x)=x^4−4x^3−8x^2+ax+b has 1+2i as a solution to f(x)=0. What is the value of 10a+b?
Answers
Answered by
Steve
another root is also 1-2i, so
(x-1)^2+4 = x^2-2x+5 is a divisor of f(x)
(x^2-2x+5)(x^2+mx+n)
= x^4 + (m-2)x^3 + (n-2m+5)x^2 + (5m-2n)x + (5n)
So,
m-2 = -4
n-2m+5 = -8
a = 5m-2n
b = 5n
m = -2
n = -17
a = 24
b = -85
10a+b = 155
check:
(x^2-2x+5)(x^2-2x-17) = x^4−4x^3−8x^2+24x-85
(x-1)^2+4 = x^2-2x+5 is a divisor of f(x)
(x^2-2x+5)(x^2+mx+n)
= x^4 + (m-2)x^3 + (n-2m+5)x^2 + (5m-2n)x + (5n)
So,
m-2 = -4
n-2m+5 = -8
a = 5m-2n
b = 5n
m = -2
n = -17
a = 24
b = -85
10a+b = 155
check:
(x^2-2x+5)(x^2-2x-17) = x^4−4x^3−8x^2+24x-85
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