The polynomial has the following zeros: 0, 1, -2i, 3+the square root of 3. Put the polynomial in standard form.
1 answer:
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)=
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)=
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Answer:
option-B
Step-by-step explanation:
we are given system of inequality as
now, we can check solution
At (0,0):
we can plug x=0 and y=0
and check inequality
so, this is FALSE
now, we can check second inequality
So, this is TRUE
so, option-B