document.write( "Question 738152: Hi, I needed help on this question...I don't know how to do it.
\n" ); document.write( "Find the solutions:\r
\n" ); document.write( "\n" ); document.write( "2x to the 4th + 8x to the 3rd = 0
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Algebra.Com's Answer #450693 by DrBeeee(684)\"\" \"About 
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I'm not sure you mean the given equation is
\n" ); document.write( "(1) (2x)^4 + (8x)^3 = 0 OR
\n" ); document.write( "(2) 2x^4 + 8x^3 = 0
\n" ); document.write( "It doesn't effect the solution technicque, only the value of one of the roots.
\n" ); document.write( "I'll do (1), where we have
\n" ); document.write( "(3) 16x^4 + 512x^3 = 0
\n" ); document.write( "Dividing by 16 gives us
\n" ); document.write( "(4) x^4 + 32x^3 = 0
\n" ); document.write( "Now factor out x^3 and get
\n" ); document.write( "(5) x^3*(x + 32) = 0
\n" ); document.write( "Since the given equation is 4th order, we need four roots,. They are given by the solution set x = {0,0,0,-32). Note that x^3 = 0 results in a triple root at x = 0.
\n" ); document.write( "If you meant that the given equation is given by (2), the solution set is {0,0,0,-4)
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