solve the given polynomial equation, find the solution set of the equation x^4-2x^3-25x^2-34x-12=0
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x4-2x3-25x2-34x-12=0.

Try rational zeroes first. So we look at the factors of 12: 1, 2, 3, 4, 6.

Start with x=1 or -1:

Clearly x=1 is going to give us a negative result, so let's try x=-1:

1+2-25+34-12=37-37=0, so x=-1, that is, x+1=0, x+1 is a factor. Divide by this factor using synthetic division by the root:

-1 | 1 -2 -25 -34  -12

      1 -1    3   22 | 12

      1 -3 -22 -12 |   0 = x3-3x2-22x-12.

Now try the next rational zero, x=3 or x=-3. x=3 clearly gives us a negative number so try x=-3:

-27-27+66-12=-66+66=0, so x=-3 is a root. Divide by the root:

-3 | 1 -3 -22  -12

      1 -3  18 | 12

      1 -6   -4 |   0 = x2-6x-4

The quadratic remains to be solved.

x2-6x-4=x2-6x+9-13=0,

(x-3)2-13=0,

x-3=±√13, x=3+√13=6.6056 or x=3-√13=-0.6056 approx.

by Top Rated User (1.0m points)

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