Simplifying: \frac{x^2-6x+4}{(x^2-4)(x-2)}

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To simplify the expression \(\frac{x^2-6x+4}{(x^2-4)(x-2)}\), first combine the fractions to form a single numerator. The numerator can be factored as \(x^2-x-2\), which simplifies to \((x-2)(x+1)\). This allows for the cancellation of the common factor \((x-2)\) in both the numerator and denominator. The final simplified expression is \(\frac{x+1}{(x-2)(x+2)}\). Understanding the rules of factoring and cancellation is key to reaching the solution.
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How do I simplify this?

\frac{1}{x-2} - \frac{1}{x^2-4} = \frac{(x^2-4)-(x-2)}{(x^2-4)(x-2)}
 
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Combine the 4 and 2 in the numerator so you have a single quadratic. Then see if it factors so you can see of if there are any common factors you can cancel.
 
\frac{x^2-x-2}{(x^2-4)(x-2)}

I know I should be able to see this instantly but I'm not! What comes next?
 
You don't have to see it instantly. Can you factor the numerator?
 
Like this?

\frac{x^2-x-2}{(x-2)(x+2)(x-2)}
 
Last edited:
\frac{x^2-x-2}{(x-2)^2(x+2)} = \frac{x+1}{(x-2)(x+2)}

The final step is the answer but I don't know how to get there. What are the rules here?
 
danni7070 said:
Like this?

\frac{x^2-x-2}{(x-2)(x+2)(x-2)}


after this stage

factor the numerator

x2 - x - 2 = (x-2)(x+1)

right?

then u can cancel t he (x-2) in the numerator and denominator and ur left with that answer that ur looking for
 
I'm having one of those dumb days right know. Jesus!

Thank you Dick and salman213.
 
danni7070 said:
Like this?

\frac{x^2-x-2}{(x-2)(x+2)(x-2)}
They said factor the numerator! (Although factoring the denominator doesn't hurt.)
 
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