document.write( "Question 624554: I need to factor the polynomials and solve the equation
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\n" ); document.write( "32xy^3 +4x=\r
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\n" ); document.write( "\n" ); document.write( "I factored out 4x to get 4x(8y^3+1)
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Algebra.Com's Answer #392841 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Now let's factor \"8y%5E3%2B1\"\r
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\n" ); document.write( "\n" ); document.write( "\"8y%5E3%2B1\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"%282y%29%5E3%2B%281%29%5E3\" Rewrite \"8y%5E3\" as \"%282y%29%5E3\". Rewrite \"1\" as \"%281%29%5E3\".\r
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\n" ); document.write( "\n" ); document.write( "\"%282y%2B1%29%28%282y%29%5E2-%282y%29%281%29%2B%281%29%5E2%29\" Now factor by using the sum of cubes formula. Remember the sum of cubes formula is \"A%5E3%2BB%5E3=%28A%2BB%29%28A%5E2-AB%2BB%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%282y%2B1%29%284y%5E2-2y%2B1%29\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "So \"8y%5E3%2B1\" factors to \"%282y%2B1%29%284y%5E2-2y%2B1%29\".\r
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\n" ); document.write( "\n" ); document.write( "In other words, \"8y%5E3%2B1=%282y%2B1%29%284y%5E2-2y%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So in the end, \"32xy%5E3+%2B4x\" completely factors to \"4x%282y%2B1%29%284y%5E2-2y%2B1%29\"\r
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\n" ); document.write( "If you need more help, email me at jim_thompson5910@hotmail.com\r
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\n" ); document.write( "\n" ); document.write( "Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you\r
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