Question 1204777
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Answer: <font color=red>f(x) = x^4  - 8x^3 + 10x^2 + 48x + 29</font>


Explanation


The polynomial has real number coefficients, so any complex root of the form a+bi has a conjugate a-bi


The root 5+2i pairs with 5-2i
x = 5+2i
x-5 = 2i
(x-5)^2 = (2i)^2
(x-5)^2 = 4i^2
(x-5)^2 = 4(-1)
(x-5)^2 = -4
(x-5)^2+4 = 0
x^2-10x+25+4 = 0
x^2-10x+29 = 0
You'll arrive at this same equation if you started with x = 5-2i


Therefore x^2-10x+29 = 0 has the complex roots x = 5+2i and x = 5-2i
The quadratic formula can be used to confirm this. 
Online CAS (computer algebra system) tools such as WolframAlpha or GeoGebra can also be used to confirm this claim.


So far we have shown that (x^2-10x+29) is a factor


If x = -1 is a root then (x+1) is a factor
This root is of multiplicity 2. It's a double root. So (x+1)^2 is a factor.


Let's expand out the following
(x+1)^2*(x^2-10x+29)
(x^2+2x+1)*(x^2-10x+29)
x^2(x^2-10x+29) + 2x(x^2-10x+29) + 1(x^2-10x+29)
(x^4-10x^3+29x^2) + (2x^3-20x^2+58x) + (x^2-10x+29)
x^4 + (-10x^3+2x^3) + (29x^2-20x^2+x^2) + (58x-10x) + 29
<font color=red>x^4  - 8x^3 + 10x^2 + 48x + 29</font>


Using a CAS is one way to confirm the answer above is correct.


Technically we could scale this polynomial up or down, but I'll leave the leading coefficient as 1.
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