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Question 1197172: Find a polynomial P(x) with real coefficients having a degree 6, leading coefficient 2, and zeros 3, 0 (multiplicity 3), and 4 - 4i.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Since P(x) has all real coefficients, this means any complex roots of the form a+bi automatically have a conjugate paired root of a-bi
More specifically for this case we have the root 4-4i pair up with 4+4i
Let's start with x = 4+4i and do a bit of algebra like so
x = 4+4i
x-4 = 4i
(x-4)^2 = (4i)^2
x^2-8x+16 = 16i^2
x^2-8x+16 = 16(-1)
x^2-8x+16 = -16
x^2-8x+16+16 = 0
x^2-8x+32 = 0
If you were to solve x^2-8x+32 = 0 using the quadratic formula, then you'd get the two roots x = 4+4i and x = 4-4i.
I'll let you do these verification steps.
The summary so far is that if 4-4i is a root of P(x), then it leads to x^2-8x+32 being a factor of P(x)
We also know that x = 3 is a root, so (x-3) is another factor of P(x)
x = 0 is a root with multiplicity 3, which tells us x^3 is a factor
We have
P(x) = 2x^3(x-3)(x^2-8x+32)
Distribute and expand everything out
P(x) = 2x^3(x-3)(x^2-8x+32)
P(x) = (2x^4-6x^3)(x^2-8x+32)
P(x) = 2x^4(x^2-8x+32)-6x^3(x^2-8x+32)
P(x) = 2x^4*x^2+2x^4*(-8x)+2x^4*(32)-6x^3(x^2)-6x^3(-8x)-6x^3(32)
P(x) = 2x^6-16x^5+64x^4-6x^5+48x^4-192x^3
P(x) = 2x^6+(-16x^5-6x^5)+(64x^4+48x^4)-192x^3
P(x) = 2x^6-22x^5+112x^4-192x^3
You can use WolframAlpha or the CAS tool in GeoGebra to confirm that we have the correct polynomial function.
Many other online tools are available.
CAS = computer algebra system
You could also graph it to look for the x intercepts, but the downside is it won't show the complex roots.
Answer: P(x) = 2x^6 - 22x^5 + 112x^4 - 192x^3
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