document.write( "Question 198237This question is from textbook algebra and trigonometry
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\n" ); document.write( "\n" ); document.write( "(4t^3-at^2-2bt+5)-(ct^3+2t^2-6t+3) = t^3-2t+d
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Algebra.Com's Answer #148688 by jim_thompson5910(35256)\"\" \"About 
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\"%284t%5E3-at%5E2-2bt%2B5%29-%28ct%5E3%2B2t%5E2-6t%2B3%29+=+t%5E3-2t%2Bd\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"4t%5E3-at%5E2-2bt%2B5-ct%5E3-2t%5E2%2B6t-3+=+t%5E3-2t%2Bd\" Distribute\r
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\n" ); document.write( "\n" ); document.write( "\"4t%5E3-at%5E2-2bt-ct%5E3-2t%5E2%2B6t-2+=+t%5E3-2t%2Bd\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"%284t%5E3-ct%5E3%29%2B%28-at%5E2-2t%5E2%29%2B%28-2bt%2B6t%29%2B%282%29+=+t%5E3-2t%2Bd\" Group like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"%284-c%29t%5E3%28-a-2%29t%5E2%2B%28-2b%2B6%29t%2B2+=+t%5E3-2t%2Bd\" Factor out the respective GCFs from each group\r
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\n" ); document.write( "\n" ); document.write( "Since the coefficient of \"t%5E3\" on the left side is \"4-c\" and on the right side is 1, this means that we know that \"4-c=1\"\r
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\n" ); document.write( "\n" ); document.write( "Because the coefficient for \"t%5E2\" on the left is \"-a-2\". However, there is no \"t%5E2\" on the right. So this means that the coefficient of \"t%5E2\" on the right is 0. So \"-a-2=0\"\r
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\n" ); document.write( "\n" ); document.write( "Finally, the coefficients for the \"t\" terms on the left and right sides are \"-2b%2B6\" and \"-2\" respectively. So this means that \"-2b%2B6=-2\"\r
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\n" ); document.write( "\n" ); document.write( "Take note that the only constants on the left and right sides are \"2\" and \"d\". So we know that \"2=d\" or \"d=2\"\r
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\n" ); document.write( "\n" ); document.write( "So all you need to do is solve the equations:\r
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\n" ); document.write( "\n" ); document.write( "\"system%284-c=1%2C-a-2=0%2C-2b%2B6=-2%29\"\r
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\n" ); document.write( "\n" ); document.write( "To find the values of 'a', 'b', and 'c'. I'll let you do that.\r
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