SOLUTION: form a polynomial f(x) with real coefficients having the given degree and zeros degree :4, zeros: 5+2i, -1 multiplicity 2

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Question 1204777: form a polynomial f(x) with real coefficients having the given degree and zeros degree :4, zeros: 5+2i, -1 multiplicity 2
Found 3 solutions by josgarithmetic, math_tutor2020, MathLover1:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
This is a very routine exercise. Almost to fitting a formula!

%28x-%285%2B2i%29%29%28x-%285-2i%29%29%28x%2B1%29%5E2
Carry through all the multiplication and simplifications....


%28x-5-2i%29%28x-5%2B2i%29%28x%2B1%29%5E2
%28%28x-5%29-2i%29%28%28x-5%29%2B2i%29%28x%5E2%2B2x%2B1%29
%28%28x-5%29%5E2-%282i%29%5E2%29%28x%5E2%2B2x%2B1%29, recognize what makes Difference Of Two Squares
%28x%5E2-10x%2B25-4i%5E2%29%28x%5E2%2B2x%2B1%29
%28x%5E2-10x%2B25-4%28-1%29%29%28x%5E2%2B2x%2B1%29, understand meaning of i^2
%28x%5E2-10x%2B29%29%28x%5E2%2B2x%2B1%29

Choosing lattice form for the polynomial multiplication
            x^2        2x        1
--------------------------------------------
x^2    |    x^4         2x^3      x^2
       |
-10x   |   -10x^3      -20x^2     -10x
       |
29     |  29x^2         58x       29

Resulting in highlight%28x%5E4-8x%5E3%2B10x%5E2%2B48x%2B29%29

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: f(x) = x^4 - 8x^3 + 10x^2 + 48x + 29

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
x^4 - 8x^3 + 10x^2 + 48x + 29

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.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
degree :4
zeros:
x1=5%2B2i, => complex zeros always come in pairs, so you also have x2=5-2i
x3=-1+ ,multiplicity 2=> x4=-1

f%28x%29=%28x-x1%29%28x-x2%29%28x-x3%29%28x-x4%29
f%28x%29=%28x-%285%2B2i%29%29%28x-%285-2i%29%29%28x-%28-1%29%29%28x-%28-1%29%29
f%28x%29=%28x-5-2i%29%28x-5%2B2i%29%28x%2B1%29%28x%2B1%29
f%28x%29=%28x%5E2+-+10x+%2B+29%29%28x%5E2+%2B+2x+%2B+1%29
f%28x%29=x%5E4+-+8x%5E3+%2B+10x%5E2+%2B+48x+%2B+29