\({2h \over h^{2} - 9} + {h \over h^{2} + 6h + 9} - {3 \over h - 3}\)
the textbook answer is \({-3(5h + 9)\over (h + 3)(h + 3)(h - 3)}, h ≠ ± 3\) how should i get it?
+128407 2h / (h^2 - 9) + h / (h^2 + 6h + 9) - 3/ ( h - 3) factor the denominators
2h / [ ( h + 3) ( h - 3) ] + h / [ ( h + 3) (h + 3) ] - 3 / (h - 3)
The common denominator is ( h + 3) ( h + 3) ( h - 3) ....so we have
2h (h + 3) + h( h -3) - 3 (h + 3) ( h + 3)
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(h + 3) ( h + 3) ( h - 3)
Simplify and note that x cannot be either 3 or -3 because both values make the denominator = 0
So we have
2h^2 + 6h + h^2 - 3h - 3 ( h^2 + 6h + 9)
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( h + 3) ( h + 3) ( h - 3)
2h^2 + 6h + h^2 - 3h - 3h^2 - 18h - 27
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(h + 3) ( h + 3) (h - 3)
-15h - 27
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(h + 3) ( h + 3) ( h - 3)
-3(5h + 9)
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(h + 3) ( h + 3) ( h - 3)
