Simplify the rational expression. a + 3 a 2 + 6 a + 9 \frac { a + 3 } { a ^ { 2 } + 6 a + 9 } a 2 + 6 a + 9 a + 3 A) 1 7 a + 3 , a ≠ − 3 7 \frac { 1 } { 7 a + 3 } , a
eq - \frac { 3 } { 7 } 7 a + 3 1 , a  = − 7 3 B) 1 a 2 + 9 , a ≠ 3 \frac { 1 } { a ^ { 2 } + 9 } , a
eq 3 a 2 + 9 1 , a  = 3 C) 1 a + 9 , a ≠ − 9 \frac { 1 } { a + 9 } , a
eq - 9 a + 9 1 , a  = − 9 D) 1 a + 3 , a ≠ − 3 \frac { 1 } { a + 3 } , a
eq - 3 a + 3 1 , a  = − 3 E) 1 3 a + 9 , a ≠ − 3 1 \frac { 1 } { 3 a + 9 } , a
eq - \frac { 3 } { 1 } 3 a + 9 1 , a  = − 1 3

Question

Simplify the rational expression.   a + 3 a 2 + 6 a + 9 \frac { a + 3 } { a ^ { 2 } + 6 a + 9 } a 2 + 6 a + 9 a + 3 ​ A)    1 7 a + 3 , a ≠ − 3 7 \frac { 1 } { 7 a + 3 } , a 
eq - \frac { 3 } { 7 } 7 a + 3 1 ​ , a  = − 7 3 ​ B)    1 a 2 + 9 , a ≠ 3 \frac { 1 } { a ^ { 2 } + 9 } , a 
eq 3 a 2 + 9 1 ​ , a  = 3 C)    1 a + 9 , a ≠ − 9 \frac { 1 } { a + 9 } , a 
eq - 9 a + 9 1 ​ , a  = − 9 D)    1 a + 3 , a ≠ − 3 \frac { 1 } { a + 3 } , a 
eq - 3 a + 3 1 ​ , a  = − 3 E)    1 3 a + 9 , a ≠ − 3 1 \frac { 1 } { 3 a + 9 } , a 
eq - \frac { 3 } { 1 } 3 a + 9 1 ​ , a  = − 1 3 ​

Angie Abgaryan · Accepted Answer

Final Answer : D, Explanation :  We can factor the denominator as $(a+3)^2$. Therefore, we have $\frac{a+3}{(a+3)^2}$. Canceling the common factor, we get $\frac{1}{a+3}$, which matches with option D.

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