I’m trying to simplify this without turning it into algebra soup:
(x^2 – 9)/(x^2 – x – 6) * (x – 2)/(x + 3)
My attempt: I factored the first fraction as (x – 3)(x + 3) / ((x – 3)(x + 2)). So the whole thing looks like [(x – 3)(x + 3)] / [(x – 3)(x + 2)] * (x – 2)/(x + 3).
Now I’m second-guessing what I’m allowed to cancel. Since everything’s multiplied, I think the (x – 3) terms can go. Can the (x + 3) on top also cancel with the (x + 3) in the other denominator, or is that cheating because it’s in a different fraction? If I do that, I end up with something like (x – 2)/(x + 2), and then my brain wants to “cancel the x” and make it -2/2 = -1, which feels wrong but weirdly tempting.
Also, do I need to keep track of x ≠ 3, -2, -3 even if those factors disappear after canceling? What’s the quick, clean way to see what cancels here without overthinking it?
Any help appreciated!
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3 Responses
Great instincts factoring first-you’re right on track. Since everything is multiplied, you can think of the whole thing as one big fraction: [(x − 3)(x + 3)(x − 2)]/[(x − 3)(x + 2)(x + 3)]. Now the canceling is clean: (x − 3) cancels, and (x + 3) cancels, because they’re common factors of the overall numerator and denominator, not just “in the same original fraction.” That leaves (x − 2)/(x + 2). Resist the urge to “cancel the x” here-x isn’t a factor of the whole top and bottom; it’s part of sums, so only entire matching factors can cancel. One more important detail: keep the original restrictions from any denominator you had along the way, so x ≠ 3, −2, −3 (even though 3 and −3 don’t show up after canceling). A nice summary rule: factor completely, write as a single product, cancel only identical factors, and note domain restrictions first so you don’t lose them. If you want a quick refresher with examples, this Khan Academy page is solid: https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/simplifying-rational-expressions/a/simplifying-rational-expressions. Want to try a similar one-say, (x^2 − 16)/(x^2 − x − 20) · (x − 4)/(x + 5)-and see what cancels?
I feel you on the algebra soup! You’re on the right track by factoring first: x^2 − 9 = (x − 3)(x + 3) and x^2 − x − 6 = (x − 3)(x + 2). So the whole product is [(x − 3)(x + 3)(x − 2)] / [(x − 3)(x + 2)(x + 3)]. Since everything is multiplied, you’re allowed to cancel common factors across the entire numerator and denominator-even if they came from different original fractions. So (x − 3) cancels, and (x + 3) cancels too, leaving (x − 2)/(x + 2). And yes, that “cancel the x” urge in (x − 2)/(x + 2) is a trap-cancellation only works on factors (things being multiplied), not on terms inside sums or differences.
One more important thing: keep the restrictions from the original denominators. Even though (x − 3) and (x + 3) cancel, the original expression was undefined at x = 3, −2, −3, so those values are still excluded. Final answer: (x − 2)/(x + 2), with x ≠ 3, −2, −3. Quick rule of thumb to avoid overthinking: factor everything, write it as one big fraction, cancel common factors, then list the x-values that would have made any original denominator zero.
Hope this helps!
You’ve factored correctly. From there, the clean way is:
1) Factor everything:
(x^2 − 9)/(x^2 − x − 6) · (x − 2)/(x + 3)
= [(x − 3)(x + 3)]/[(x − 3)(x + 2)] · (x − 2)/(x + 3).
2) Combine as one big fraction (multiplication lets you do this), then cancel common factors that appear as whole factors in numerator and denominator:
Overall numerator: (x − 3)(x + 3)(x − 2)
Overall denominator: (x − 3)(x + 2)(x + 3)
Now cancel (x − 3) and (x + 3). That’s allowed because they are full factors and everything is multiplied. This leaves:
(x − 2)/(x + 2).
3) Domain restrictions: even though (x − 3) and (x + 3) canceled, they were in denominators originally, so x cannot be 3, −2, or −3:
– x ≠ 3 (makes x − 3 = 0 in the first denominator),
– x ≠ −2 (makes x + 2 = 0 in the first denominator),
– x ≠ −3 (makes x + 3 = 0 in the second denominator).
So the simplified form is (x − 2)/(x + 2), with x ≠ 3, −2, −3.
Why you can’t “cancel the x” in (x − 2)/(x + 2):
– Cancellation works only with common factors, not with terms inside sums or differences. x is not a factor of (x − 2) or (x + 2) as a whole. Writing (x − 2)/(x + 2) = (x/x) + (−2/2) is not valid algebra.
Quick check with a number (simple worked example):
Take x = 4 (allowed). Original:
(16 − 9)/(16 − 4 − 6) · (4 − 2)/(4 + 3) = 7/6 · 2/7 = 1/3.
Simplified:
(4 − 2)/(4 + 2) = 2/6 = 1/3.
Matches, as it should.
Summary:
– Yes, cancel (x − 3) and (x + 3) across the product.
– Final answer: (x − 2)/(x + 2), with x ≠ 3, −2, −3.
– Don’t cancel “x” inside (x − 2)/(x + 2); that’s not a common factor.