According to the Rational Root Theorem, the possible rational roots of the polynomial x4+4x3−5x2−16x+4=0x^{4}+4x^{3}-5x^{2}-16x+4=0x4+4x3−5x2−16x+4=0 are: ±1,±2,±4 \pm 1, \pm 2, \pm 4 ±1,±2,±4
We tested the possible rational roots using synthetic division:
Testing x=1x = 1x=1:
Testing x=−1x = -1x=−1:
Testing x=2x = 2x=2:
From the tests, we found that one rational root of the given equation is: x=2 x = 2 x=2
Using the root x=2x = 2x=2, we performed polynomial long division on x4+4x3−5x2−16x+4x^{4}+4x^{3}-5x^{2}-16x+4x4+4x3−5x2−16x+4 by x−2x - 2x−2. The result of the division is: x4+4x3−5x2−16x+4x−2=x3+6x2+7x−2 \frac{x^{4} + 4x^{3} - 5x^{2} - 16x + 4}{x - 2} = x^{3} + 6x^{2} + 7x - 2 x−2x4+4x3−5x2−16x+4=x3+6x2+7x−2 with a remainder of 000.
The possible rational roots are ±1,±2,±4\pm 1, \pm 2, \pm 4±1,±2,±4. One rational root of the given equation is 222. The polynomial can be factored as: x4+4x3−5x2−16x+4=(x−2)(x3+6x2+7x−2) x^{4}+4x^{3}-5x^{2}-16x+4 = (x - 2)(x^{3} + 6x^{2} + 7x - 2) x4+4x3−5x2−16x+4=(x−2)(x3+6x2+7x−2) Thus, the final answer is: x=2 \boxed{x = 2} x=2
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