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rizzamella rizzamella
wrote...
Posts: 73
Rep: 1 0
11 years ago
These two problems are confusing me. I'm supposed to get a third-degree polynomial equation with rational coefficients that has the given numbers as roots

1. 3, 2-i

2. -4, 4i
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michquelldmichquelld
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Posts: 21
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11 years ago
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wrote...
11 years ago
1.  If 2-i is a root, then 2+i is also a root.
Use the factor theorem to convert the three roots into factors:
(x - 3) (x - 2 + i) (x - 2 - i)
Multiply these roots together:
x^3 - 7x^2 + 17x - 15

2. If 4i is a root, then -4i is also a root.
Use the factor theorem to convert the three roots into factors:
(x + 4) (x - 4i) (x + 4i)
Multiply these roots together:
x^3 + 4x^2 + 16x + 64
wrote...
11 years ago
If a real polynomial (a polynomial with coefficients that are real numbers), has a complex root, then it MUST have another complex root, and this other complex root MUST be the conjugate of the first one.

The conjugate of the complex number (a + bi) is (a - bi)
the sign in front of the imaginary part changes.

In the first problem, they give you a root as being:
2 - i
and your are told it is a real polynomial.
Therefore, you know that there MUST be another complex root and it MUST be the conjugate of 2 - i

The three roots are
z = 3
z = 2 - i
z = 2 + i

Once you have the roots, then each one corresponds to a factor:
if z = +k is a root, then
(z - k) is a factor

Given the three roots, you get
p(z) = a(z - 3)(z - 2 + i)(z - 2 - i)
where a is a real number.
If you are not told what the leading coefficient (a) should be, then the easiest choice is a=1, so that you can ignore it

p(z) = (z - 3)(z - 2 + i)(z - 2 - i)
p(z) = (z - 3)(z^2 - 4z + 5)
p(z) = z^3 - 7z^2 + 17z  - 15)
(or something like that)

In the second problem, rewrite the given complex root as
0 + 4i
then find the conjugate
(after that, you can drop the 0 again)

---

In real polynomials of ANY degree, complex roots MUST come in conjugate pairs.

If the polynomial has an odd number of degrees (like in your problems) then at least ONE root must be real.
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