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\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?

Michael Hardy
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user1068636
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3 Answers3

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HINT:

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

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

Mark Viola
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$2x^2 -x -1$ 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
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V Shreyas
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I cannot comment, so I shall answer instead.

The first thing I tend to do with quadratics like is, is rather than use the quadratic formula, I would find the discriminant. If this is a square number, then you know your quadratic will factorise nicely. You can then either do quick trial and error, or use the formula to find the solutions and deduce the factorisation.

So in this case, b^2-4ac = 9, so we know it factorises. You know it will be (2x+a)(x+b), and it is then easy to finish from here.

Hope this helps :)

Ali
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