Simplifying a rational expression by factoring numerator and denominator

Key takeaway: This problem checks whether you can factor polynomials correctly, cancel only common factors, and rewrite the rational expression in simplest form.

Factor numerator and denominator

We start with

$\frac{x^{2} - 2x}{x^{2} - x - 2}$.

Factor the numerator:

$x^{2} - 2x = x(x - 2)$.

Factor the denominator:

$x^{2} - x - 2 = (x - 2)(x + 1)$.

Cancel the common factor

Now the expression becomes

$\frac{x(x - 2)}{(x - 2)(x + 1)}$.

Since $(x - 2)$ appears in both numerator and denominator, it cancels, leaving

$\frac{x}{x + 1}$.

So the simplest form is

$\boxed{\frac{x}{x+1}}$, which is choice (c).

Domain restriction matters

Even after simplifying, the original denominator cannot be zero. So the original expression is undefined when

$x - 2 = 0$ or $x + 1 = 0$,

which means $x \neq 2$ and $x \neq -1$.

Common simplification rule

Only factors can be canceled, not terms. That is why you must factor first before reducing the expression.

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Pitfalls the pros know 👇 A frequent mistake is canceling the $x$ terms directly, as if $\frac{x^{2}-2x}{x^{2}-x-2}$ were $\frac{x^{2}}{x^{2}}$ or $\frac{-2x}{-x-2}$. Cancellation works only on common factors, not on pieces of sums or differences. Another error is forgetting the restrictions on $x$ after simplification; the simplified form is not valid at values that made the original denominator zero.

What if the problem changes? If the expression were $\frac{x^{2} - 4x}{x^{2} - 3x - 4}$, you would again factor numerator and denominator first. That becomes $\frac{x(x-4)}{(x-4)(x+1)}$, which simplifies to $\frac{x}{x+1}$ with the restriction $x \neq 4,-1$. If the numerator and denominator shared no factor, the expression would stay unchanged except for factorization and domain notes.

Tags: rational expression, common factor cancellation, domain restriction