document.write( "Question 235807: can someone help me pls\r
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Algebra.Com's Answer #173610 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
Your equation is:\r
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\n" ); document.write( "\n" ); document.write( "x^3-4x^2-5x+20\r
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\n" ); document.write( "\n" ); document.write( "divide this equation by (x-4) to get:\r
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\n" ); document.write( "\n" ); document.write( "(x-4) * (x^2-5)\r
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\n" ); document.write( "\n" ); document.write( "This factors out to:\r
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\n" ); document.write( "\n" ); document.write( "(x-4) * (x-sqrt(5)) * (x+sqrt(5))\r
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\n" ); document.write( "\n" ); document.write( "I confirmed it's accurate by replacing x with 5 in both the original equation and the factored equation to get the same answer.\r
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\n" ); document.write( "\n" ); document.write( "You can also multiply the factors out again to get back to the original equation.\r
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\n" ); document.write( "\n" ); document.write( "I figured (x-4) might divide into the original equation because -4 * x^2 = -4x^2 which would subtract nicely from -4x^2 in the original equation, and becausse -4 * -5 = 20 which would subtract nicely from 20 at the end of the equation.\r
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\n" ); document.write( "\n" ); document.write( "Not totally scientific but it worked.\r
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\n" ); document.write( "\n" ); document.write( "x goes into x^3 by a factor of x^2\r
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\n" ); document.write( "\n" ); document.write( "(x-4) * x^2 = x^3 - 4x^2 which subtracts from x^3 - 4x^2 without a remainder.\r
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\n" ); document.write( "\n" ); document.write( "The division is left with (x-4) into (-5x + 20) which comes out very clean with a factor of -5.\r
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\n" ); document.write( "\n" ); document.write( "-5 * (x-4) = -5x + 20 which subtracts very nicely from -5x + 20 to leave a remainder of 0.\r
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\n" ); document.write( "\n" ); document.write( "My answer by dividing the original equation by (x-4) became (x-4) * (x^2 - 5) which led to the final solution.\r
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