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Given that -6i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable.

f(x)=x^4-15x³ + 90x² - 540x + 1944

User JayantS
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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.

User Angelo Mendes
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