What is the left overs of 185 devided by 6
1 answer:
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Answer:
One other zero is 2+3i
Step-by-step explanation:
If 2-3i is a zero and all the coefficients of the polynomial function is real, then the conjugate of 2-3i is also a zero.
The conjugate of (a+b) is (a-b).
The conjugate of (a-b) is (a+b).
The conjugate of (2-3i) is (2+3i) so 2+3i is also a zero.
Ok so we have two zeros 2-3i and 2+3i.
This means that (x-(2-3i)) and (x-(2+3i)) are factors of the given polynomial.
I'm going to find the product of these factors (x-(2-3i)) and (x-(2+3i)).
(x-(2-3i))(x-(2+3i))
Foil!
First: x(x)=x^2
Outer: x*-(2+3i)=-x(2+3i)
Inner: -(2-3i)(x)=-x(2-3i)
Last: (2-3i)(2+3i)=4-9i^2 (You can just do first and last when multiplying conjugates)
---------------------------------Add together:
x^2 + -x(2+3i) + -x(2-3i) + (4-9i^2)
Simplifying:
x^2-2x-3ix-2x+3ix+4+9 (since i^2=-1)
x^2-4x+13 (since -3ix+3ix=0)
So x^2-4x+13 is a factor of the given polynomial.
I'm going to do long division to find another factor.
Hopefully we get a remainder of 0 because we are saying it is a factor of the given polynomial.
x^2+1
---------------------------------------
x^2-4x+13| x^4-4x^3+14x^2-4x+13
-( x^4-4x^3+ 13x^2)
------------------------------------------
x^2-4x+13
-(x^2-4x+13)
-----------------
0
So the other factor is x^2+1.
To find the zeros of x^2+1, you set x^2+1 to 0 and solve for x.
So the zeros are i, -i , 2-3i , 2+3i