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Find the zeros of f. (Enter your answers as a comma-separated list.) (hold on while I copy the equation ><)
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\[x ^{2}\left( 2e ^{3x} \right)+8xe ^{3x}+31e ^{3x}+8xe ^{3x}\]
hello the first step to this problem is to factor out 2e^3x can you do this for me and put it here?
Okay
x^2+8x+31
don't you mean 16x? there seems to be 2 8x's
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If that works I /might/ be able to solve it. I didn't think of facotring out 2e^3x And there are, so combine them and you get 16x, but divide them by 2 and it becomes 8, no? In that case the 31 should be 31/2
YES! I got the answer! Thank you so much. :)
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