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Novosadov [1.4K]
3 years ago
7

15 Points!

Mathematics
1 answer:
Virty [35]3 years ago
6 0
5; -3; -1 -2i
<span>In polynomials we have not of an imaginary unit (i), therefore we have the fourth zeros -1-2i.</span>

f(x)=(x-5)(x+3)\left[x-(-1-2i)\right]\left[x-(-1+2i)\right]\\\\=(x^2+3x-5x-15)[x^2-x(-1+2i)-x(-1-2i)+(-1)^2-(2i)^2]\\\\=(x^2-2x-15)(x^2+x-2i+x+2i+1+4)\\\\=(x^2-2x-15)(x^2+2x+5)\\\\=x^4+2x^3+5x^2-2x^3-4x^2-10x-15x^2-30x-75\\\\=x^4-14x^2-40x-75

Answer: B)


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