Question 76465
Check to make sure that you copied the problem correctly. If the problem is correct as
you posted it, the answer is {{{5x^2 + 15x + 40 +(121/(x-3))}}}
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If you posted the problem correctly, then I suspect that you are adding not subtracting 
during the division process.
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For example.  You got the first division correct ... and the first quotient term is, as you
found, {{{5x^2}}}. When you multiply this back times the divisor {{{x-3}}} the product is:
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{{{5x^3 - 15x^2}}}
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when you subtract this from {{{5x^3 + 0x^2}}} you CHANGE THE SIGNS of {{{5x^3 - 15x^2}}} to
get {{{-5x^3 + 15x^2 }}} and then you add this to {{{5x^3 + 0x^2}}}.  The result is:
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{{{+15x^2}}}.
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You then bring down the -5x term.  So the next division is {{{x-3}}} into {{{15x^2 - 5x}}}.
This division will result in +15x and multiplying it back the {{{+15x}}} times {{{x-3}}}
results in {{{15x^2 - 45x}}}. Subtract this from {{{15x^2 - 5x}}}.  Do that by changing the
signs to {{{-15x^2 +45x}}} and adding it to {{{15x^2 - 5x}}} to get {{{+40x}}}.
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Divide the {{{40x}}} by {{{x-3}}} to get 40.  Back multiply this to get {{{40*(x-3)}}}
equals {{{40x - 120}}}. Subtract this from {{{40x + 1}}} by changing the signs to 
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{{{-40x + 120}}} and adding that to {{{40x+1}}} to get {{{121}}}.  The 121 is the remainder
and it can be divided by the divisor {{{(x-3)}}}. This will make the answer to this division
problem:
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. {{{5x^2 + 15x + 40}}} with a remainder of {{{121/(x-3)}}}
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Hope this helps you to understand the process of algebraic long division.
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