document.write( "Question 663778: Factor the polynomial? 2x^3+4x^2+8x \n" ); document.write( "
Algebra.Com's Answer #412940 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"2x%5E3%2B4x%5E2%2B8x\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"2x%28x%5E2%2B2x%2B4%29\" Factor out the GCF \"2x\".\r
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\n" ); document.write( "\n" ); document.write( "Now let's try to factor the inner expression \"x%5E2%2B2x%2B4\"\r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression \"x%5E2%2B2x%2B4\", we can see that the first coefficient is \"1\", the second coefficient is \"2\", and the last term is \"4\".\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient \"1\" by the last term \"4\" to get \"%281%29%284%29=4\".\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to \"4\" (the previous product) and add to the second coefficient \"2\"?\r
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\n" ); document.write( "\n" ); document.write( "To find these two numbers, we need to list all of the factors of \"4\" (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of \"4\":\r
\n" ); document.write( "\n" ); document.write( "1,2,4\r
\n" ); document.write( "\n" ); document.write( "-1,-2,-4\r
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\n" ); document.write( "\n" ); document.write( "Note: list the negative of each factor. This will allow us to find all possible combinations.\r
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\n" ); document.write( "\n" ); document.write( "These factors pair up and multiply to \"4\".\r
\n" ); document.write( "\n" ); document.write( "1*4 = 4
\n" ); document.write( "2*2 = 4
\n" ); document.write( "(-1)*(-4) = 4
\n" ); document.write( "(-2)*(-2) = 4\r
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\n" ); document.write( "\n" ); document.write( "Now let's add up each pair of factors to see if one pair adds to the middle coefficient \"2\":\r
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First NumberSecond NumberSum
141+4=5
222+2=4
-1-4-1+(-4)=-5
-2-2-2+(-2)=-4
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\n" ); document.write( "\n" ); document.write( "From the table, we can see that there are no pairs of numbers which add to \"2\". So \"x%5E2%2B2x%2B4\" cannot be factored.\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So \"2x%5E3%2B4x%5E2%2B8x\" simply factors to \"2x%28x%5E2%2B2x%2B4%29\"\r
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\n" ); document.write( "\n" ); document.write( "In other words, \"2x%5E3%2B4x%5E2%2B8x=2x%28x%5E2%2B2x%2B4%29\".
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