Start with the answer format we want, and work your way toward forming a single fraction like so
[tex]a + \frac{b}{x+2}\\\\a*1+\frac{b}{x+2}\\\\a*\frac{x+2}{x+2}+\frac{b}{x+2}\\\\\frac{a(x+2)}{x+2}+\frac{b}{x+2}\\\\\frac{a(x+2)+b}{x+2}\\\\\frac{ax+2a+b}{x+2}\\\\\frac{ax+(2a+b)}{x+2}\\\\[/tex]
Compare that last expression to (2x+1)/(x+2). Notice how the ax and 2x match up, so a = 2 must be the case.
Then we have 2a+b as the remaining portion in the numerator. Plugging in a = 2 leads to 2a+b = 2*2+b = 4+b. Set this equal to the +1 found in (2x+1)/(x+2) to have the terms match.
So, 4+b = 1 leads to b = -3
Therefore, a = 2 and b = -3
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An alternative route:
[tex]\frac{2x+1}{x+2}\\\\\frac{2x+1+0}{x+2}\\\\\frac{2x+1+4-4}{x+2}\\\\\frac{(2x+4)+1-4}{x+2}\\\\\frac{2(x+2)-3}{x+2}\\\\\frac{2(x+2)}{x+2}+\frac{-3}{x+2}\\\\2-\frac{3}{x+2}\\\\[/tex]
I added and subtracted 4 in the third step so that I could form 2x+4, which then factors to 2(x+2). That way I could cancel out a pair of (x+2) terms toward the very end.
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Other alternative methods involve synthetic division or polynomial long division. They are slightly separate but related concepts.
