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saul85 [17]
3 years ago
15

The polynomial has the following zeros: 0, 1, -2i, 3+the square root of 3. Put the polynomial in standard form.

Mathematics
1 answer:
dimulka [17.4K]3 years ago
6 0
Standard form
ok, so the factored form of a polynomial with roots r1,r2,r3,r4 is
f(x)=(x-r1)(x-r2)(x-r3)(x-r4)
so
since the roots are 0,1,-2i,3+√3
I am assuming you want real coefients so ince -2i is a root, 2i is also a root

f(x)=(x-0)(x-1)(x+2i)(x-2i)(x-(3+√3))
f(x)=x⁵-(√3)x⁴-4x⁴+(√3)x³+7x³-(4√3)x²-16x²+12x

if you were allowed to have no-real coefients then exclue the 2i
f(x)=(x-0)(x-1)(x+2i)(x-(3+√3))
f(x)=(4x^4-4x^3- \sqrt{3}x^3+\sqrt{3}x^2+3x^2)+(2x^3-8x^2-2 \sqrt{3}x^2+6x+2 \sqrt{3}x) i
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Answer:

\large\boxed{a_n=3n-10}

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