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Expand the brackets:


2x⁴+8x³+60x²+104x≥0, x⁴+4x³+30x²+52x≥0, x(x³+4x²+30x+52)≥0, x(x+2)(x²+2x+26)≥0.

Since the quadratic has only complex zeroes, we have x=0 and x=-2 as the real zeroes. To satisfy the inequality, we have x≥0 or x≤-2.


82 is close to 81 which is 3⁴. So if we can make the contents of one pair of brackets=±3 and the other to ±1 we have solved the problem. So x+3=3 and x-1 gives us 3 when x=0 and 1 and -3 when x=-2. This gives us the zeroes. To find the solution for the inequality we can see that x≥0 and x≤-2 satisfies the inequality.

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