Algebra -> Polynomials-and-rational-expressions -> SOLUTION: Write a polynomial fx that satisfies the given conditions. Degree 3 polynomial with integer coefficients with zeros −5i and 6/5 Log On
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- degree\:(x+3)^{3}-12
- degree\:57y-y^{2}+(y+1)^{2}
- degree\:(2x+3)^{3}-4x^{3}
- degree\:3x+8x^{2}-4(x^{2}-1)
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- -4x^3+6x^2+2x=0
- 6+11x+6x^2+x^3=0
- 2x^5+x^4-2x-1=0
- 11+6x+x^2=-\frac{6}{x}
- x^3-2x=0
- 2x^5+x^4-2x-1=0
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Nadya C.
2 Answers By Expert Tutors
First, find the factors. For one complex zero, you also have to have the complex conjugate as another zero. So 2i would be the 3rd zero.
(x+2i)(x-2i)(5x-8) Now multiply. Will FOIL the first 2 terms.
(x2-2i+2i-4i2)(5x-8) Simplify
(x2+4)(5x-8) Now FOIL again
f(x) = 5x3-8x2+20x-32
John C. answered • 04/09/20
Experienced High-Level Math Teacher and Tutor
If x = -2i is a zero, then x = 2i is also a zero. Thus, our three zeros are:
x = - 2i , x = 2i and x = 8/5
Now, if we get them into factor form, we have:
x + 2i = 0, x - 2i = 0 and x - 8/5 = 0
(We multiply both sides of the last factor by 5 to get rid of the fraction)
x + 2i = 0, x - 2i = 0 and 5x - 8 = 0
Thus, we can start creating our polynomial:
f(x) = a(x + 2i)(x - 2i)(5x - 8)
We can let a = 1 for simplicity and start foiling the factors:
f(x) = a(x + 2i)(x - 2i)(5x - 8)
f(x) = (x + 2i)(x - 2i)(5x - 8)
f(x) = (x2 + 4)(5x - 8)
f(x) = 5x3 - 8x2 + 20x - 32
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