Question 1192627: Find a polynomial function P of the lowest possible degree, having real coefficients, a leading coefficient of 1, and with the given zeros.
2+2i, -1, and 2
any help is very appreciated
Found 2 solutions by josgarithmetic, math_tutor2020:
Answer by josgarithmetic(39618)   (Show Source):
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
The term "root" is the same as a "zero" of a function.
x = 2+2i is one root
x = 2-2i is the conjugate pair to the previous root
The roots come in conjugate pairs like this to ensure that the coefficients of P(x) are real numbers.
The general template is that a+bi and a-bi are complex conjugate pairs.
Let's apply these steps
x = 2+2i
x-2 = 2i
(x-2)^2 = (2i)^2
(x-2)^2 = 4i^2
(x-2)^2 = 4(-1)
(x-2)^2 = -4
(x-2)^2+4 = 0
You should find that x = 2-2i also leads to the equation above.
Expand things out and simplify.
(x-2)^2+4 = 0
x^2-4x+4+4 = 0
x^2-4x+8 = 0
To verify things so far, use the quadratic formula for x^2-4x+8 and you should get the complex roots x = 2+2i and x = 2-2i.
So far we've shown that if x = 2+2i and x = 2-2i are roots of P(x), then x^2-4x+8 is a factor of P(x).
Similarly, if x = -1 is a root, then x+1 is a factor.
And if x = 2 is a root, then x-2 is another factor.
The three factors we have are:
x^2-4x+8
x+1
x-2
Multiply them out and expand like so
(x^2-4x+8)(x+1)(x-2)
(x^2-4x+8)(x^2-2x+1x-2)
(x^2-4x+8)(x^2-x-2)
x^2(x^2-x-2)-4x(x^2-x-2)+8(x^2-x-2)
x^4-x^3-2x^2-4x^3+4x^2+8x+8x^2-8x-16
x^4-5x^3+10x^2-16
Therefore, the polynomial
P(x) = x^4-5x^3+10x^2-16
has the roots of x = 2-2i, x = 2+2i, x = -1, x = 2
This can be confirmed using a tool like WolframAlpha.
Answer: P(x) = x^4-5x^3+10x^2-16
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