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July 31, 2014

July 31, 2014

Posted by **Breana** on Thursday, January 13, 2011 at 12:41pm.

Y=x^4-6x^3+9x^2-6x+8

- Precal -
**Henry**, Friday, January 14, 2011 at 8:51pmY = X^4 - 6X^3 + 9X^2 -6X + 8 = 0,

It was determined by trial and error

that when X = 2, Y = 0. Therefore, 2

is a real solution.

X = 2,

X - 2 = 0,

Using synthetic division, divide the

4th degree Eq by X - 2 and get:

X^3 -4X^2 + X - 4.

Now we have:

Y = (X - 2)(X^3 - 4X^2 + X - 4 = 0,

In the cubic Eq,when X = 4, Y = 0:

X = 4,

X - 4 = 0.

We divide the cubic Eq by X - 4 and

get:

X^2 + 1.

The factored form of our 4th degree Eq is:

Y = (X - 4)(X - 2)(X^2 + 1) = 0.

X - 4 = 0,

X = 4.

X - 2 = 0,

X = 2.

X^2 + 1 = 0,

X^2 = -1,

X = sqrt(-1) = i.

Solution Set: X = 4, X = 2, X = i.

So we have 2 real and 1 imaginary sol.

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