document.write( "Question 1198625: Write a polynomial division that has a quotient of x+5 and a remainder -2
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Algebra.Com's Answer #832243 by math_tutor2020(3820)\"\" \"About 
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\n" ); document.write( "n = numerator
\n" ); document.write( "d = denominator
\n" ); document.write( "q = quotient = x+5
\n" ); document.write( "r = -2 = remainder\r
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\n" ); document.write( "\n" ); document.write( "n/d = q + (r/d)
\n" ); document.write( "n = d*q + r
\n" ); document.write( "n = d*(x+5) + (-2)
\n" ); document.write( "n = d*(x+5) - 2\r
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\n" ); document.write( "\n" ); document.write( "Now pick anything you like for the denominator d.
\n" ); document.write( "I find its easiest to select a binomial.
\n" ); document.write( "I'll go for d = x+1.\r
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\n" ); document.write( "\n" ); document.write( "n = d*(x+5) - 2
\n" ); document.write( "n = (x+1)*(x+5) - 2
\n" ); document.write( "n = (x^2+5x+1x+5) - 2
\n" ); document.write( "n = x^2+6x+3\r
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\n" ); document.write( "\n" ); document.write( "Therefore, dividing (x^2+6x+3)/(x+1) will result in a quotient of x+5 and remainder -2.
\n" ); document.write( "I'll let you use either polynomial long division or synthetic division to confirm this claim.\r
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\n" ); document.write( "\n" ); document.write( "Infinitely many rational functions of the form n/d will lead to q = x+5 and r = -2.
\n" ); document.write( "In other words, (x^2+6x+3)/(x+1) isn't the only possibility here.
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