Precal
posted by Breana .
What are the roots, real and imaginary of:
Y=x^46x^3+9x^26x+8

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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