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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?

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    $\begingroup$ $2x^2-x-1=2(x+1/2)(x-1)=(2x+1)(x-1)$ $\endgroup$ – Oussama Boussif Sep 2 '15 at 21:46
  • $\begingroup$ 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. $\endgroup$ – user249332 Sep 2 '15 at 21:48
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    $\begingroup$ Try factoring $2x^2-x-1$ again. There is an error. Afterwards, note that $2x+1=2(x+\frac{1}{2})$ $\endgroup$ – John Joy Sep 2 '15 at 21:54
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HINT:

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

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

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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 :

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

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

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