use the given zeros to find the remaining zeros of the function

h(x)=x^4-9x^3+12x^2+138x-1360
zero:3-5i

enter the remaining zeros of f

please show work

(3 - 5i) and (3+5i)

therefore divide by product(x-3+5i) and (x-3-5i)
or
(x^2 - 6 x + 34)
so do long division

x^4-9x^3+12x^2+138x-1360
-------------------------
x^2 - 6 x + 34

which is (x^2-3x-40) whew, no remainder

find the two zeros of that

x = [ 3 +/- sqrt (9+160)]/2
= [3+/-13]/2
= -5 and 8

so in the end
-5
+8
-3-5i
-3+5i

That was fun :)

-5

+8
3-5i
3+5i

To find the remaining zeros of the function, we can use the fact that complex zeros occur in conjugate pairs. Since the zero given is 3-5i, we know that its conjugate is also a zero: 3+5i.

Now let's use synthetic division to divide the polynomial by (x - 3 - 5i). Remember that if a polynomial is divided by a factor (x - a), the remainder is zero, so we can use synthetic division to check if (x - 3 - 5i) is a factor of h(x).

1. Set up the synthetic division table:

3 + 5i | 1 -9 12 138 -1360
|_________

2. Start by bringing down the coefficient of the highest power term, which is 1:

3 + 5i | 1 -9 12 138 -1360
_______
1

3. Multiply (3 + 5i) by 1 and write the result under the next coefficient:

3 + 5i | 1 -9 12 138 -1360
_______
1
-6 + 10i

4. Add the numbers below:

3 + 5i | 1 -9 12 138 -1360
_______
1
-6 + 10i
________
1 -6 + 10i

5. Multiply (3 + 5i) by -6 + 10i and write the result under the next coefficient:

3 + 5i | 1 -9 12 138 -1360
_______
1
-6 + 10i
________
1 -6 + 10i
0 0 -6 + 10i

6. Add the numbers below:

3 + 5i | 1 -9 12 138 -1360
_______
1
-6 + 10i
________
1 -6 + 10i
0 0 -6 + 10i
________
1 -6 + 10i 132 + 10i

Since the remainder is not zero, (x - 3 - 5i) is not a factor of h(x).

This means that 3-5i must be a double zero of the function, and the remaining zeros must be its conjugate: 3+5i. So we have found all the zeros of the function: 3-5i, 3+5i.

Therefore, the remaining zeros of the function h(x) are 3+5i and 3-5i.