After factoring out $x$ from the numerator, check whether the quadratic is a perfect square trinomial. Also, notice that the denominator $x^2 - 9$ has the form $a^2 - b^2$.

Once both numerator and denominator are fully factored, identify any common factors you can cancel. Remember you can only cancel factors (multipliers), not terms in a sum or difference, and you must avoid values of $x$ that make the denominator zero.

In Desmos, type the original function as

f(x) = (x^3 - 6x^2 + 9x)/(x^2 - 9)

and note that it will be undefined at $x = -3$ and $x = 3$ (there will be holes or breaks there).

Type each choice as its own function, for example:

• g(x) = x(x-3)/(x+3)
• h(x) = x(x+3)/(x-3)
• p(x) = (x-3)/(x+3)
• q(x) = x/(x+3)

Make sure you use parentheses exactly as written.

Either:

• Turn on the graphs and see which choice has the same graph as $f(x)$ everywhere that $f(x)$ is defined (ignoring the points where $f(x)$ is undefined), or
• Use a table (click the gear next to each function, then "Table") and compare values of $f(x)$ with each choice at several $x$ values like $x = 0, 1, 2, 4$. The correct choice will always match $f(x)$ at these values, while the others will not.

Start with the numerator:

First, factor out the greatest common factor $x$:

Now factor the quadratic $x^2 - 6x + 9$. Notice it is a perfect square trinomial:

• $x^2$ is $(x)^2$
• $9$ is $3^2$
• The middle term $-6x$ is $2 \cdot x \cdot (-3)$

So:

Thus the numerator becomes:

Now factor the denominator:

This is a difference of squares:

• $x^2 = x^2$
• $9 = 3^2$

So:

Substitute the factored forms of the numerator and denominator into the original fraction:

Now the expression is fully factored, so we can see common factors in the numerator and denominator.

The numerator and denominator both contain a factor of $(x - 3)$.

For $x \neq 3$ and $x \neq -3$ (the values that make the denominator zero), we can cancel one factor of $(x - 3)$:

So, for all real numbers $x$ where the original expression is defined (that is, $x \neq -3$ and $x \neq 3$), the expression is equivalent to

This matches choice A.