Answer:
Explanation:
The Conjugate Roots Theorem states that imaginary roots come in pairs. If -8i is a root of the polynomial, then so is +8i.
If both -8i and +8i are factors, we rewrite them as (x + 8i) and (x - 8i) in factored form. Since imaginary terms aren't accepted, you can FOIL these to get a quadratic with no imaginary terms.
x2 - 8ix + 8ix - 64i2
Simplify. Replace i2 with -1 because √-1 = i
x2 - 64(-1)
x2 + 64
The quadratic x2 + 64 has roots -8i and 8i. You can use the quadratic formula to verify this.
Now, let's tackle the rest of the polynomial. If x2 + 64 is a factor of the polynomial
x4 + 10x3 + 85x2 + 640x + 1344, divide it out to see what is left. I will use long division of polynomials. Note, I am using the square root symbol as a division symbol.
x2 + 10x + 21
x2 + 64 √ (x4 + 10x3 + 85x2 + 640x + 1344)
x4 + 64x2
_____________
10x3 + 21x2
10x3 + 640x
__________
21x2 +1344
21x2 + 1344
___________
0
There is no remainder, meaning x2 + 64 went in evenly. The quotient left when we divide out x2 + 64 from
x4 + 10x3 + 85x2 + 640x + 1344 is x2 + 10x + 21. Factor the quotient to get (x + 7)(x + 3).
x4 + 10x3 + 85x2 + 640x + 1344 = (x2 + 64)(x + 7)(x + 3)
You can check this factored form by FOILing out, and you should get the original polynomial.