Precal

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What are the roots, real and imaginary of:

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

• Precal -

Y = 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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