How do you factor $\frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}=$?

\begin{align} & \frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}= \frac{2x^2-x-1}{(x-3)(x+3)} \cdot \frac{x+3}{2x+1} \\[10pt] = {} & \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)(2x+1)} \end{align}

Then I use quadratic formula on numerator to factor it :

$a=2,b=-1,c=-1$

$$=\frac{2(x+2)(x-\frac{5}{2})}{(x-3)(2x+1)}$$

But apparently this can be factored further. What else can I do?

• $2x^2-x-1=2(x+1/2)(x-1)=(2x+1)(x-1)$ – Oussama Boussif Sep 2 '15 at 21:46
• You better try this yourself and increase your intuition. $2x^2-x-1=(2x+1)(x-1)$ and $x^2-9=(x+3)(x-3)$. it's simple. – user249332 Sep 2 '15 at 21:48
• Try factoring $2x^2-x-1$ again. There is an error. Afterwards, note that $2x+1=2(x+\frac{1}{2})$ – John Joy Sep 2 '15 at 21:54

HINT:

$2x^2-x-1=(x-1)(2x+1)$

$x^2-9=(x+3)(x-3)$

$2x^2 -x -1$ has $2$ factors: $x-1$ and $x+1/2$ . Hence: $$2x^2 -2x -1 = (2x+1)(x-1)$$ You have done the factorization wrong.

Hence, final answer will be :

$$(x-1)/(x-3)$$

• @Oussama Boussif - Thanks. That was my problem. I didn't apply quadratic formula correctly. – user1068636 Sep 2 '15 at 21:56
• @user1068636 I think you want to thank V Shreyas for writing the answer. But thanks – Oussama Boussif Sep 2 '15 at 22:00