# Does factoring change the result? [closed]

If $$\frac{x^2-25}{x-5}$$ equals to $$x+5$$ (after factoring),why when 5 is plugged_in the result is not the same? why is that? does factoring effect the end result ?

• I feel like this is something you easily could have Googled. We expect users to show an attempt to resolve their questions on their own. Look up “removable discontinuity.” – gen-z ready to perish May 23 at 2:14

When you factor and cancel, you are making a key assumption: that the denominator is not zero (we make this assumption every time we do division!). In other words, $$\frac{x^2 - 25}{x - 5} = \frac{(x+5)(x-5)}{(x-5)} = x+5$$ is only valid if $$x\neq 5$$ to begin with.

• i wrote the question wrong ,sorry about that – steve May 23 at 1:33
• With the new question, one of the new factors and the result should be $x+5$. That does not change the fact that the approach is correct. – Ross Millikan May 23 at 1:39
• Edited to reflect updated question. – vanPelt2 May 23 at 2:45

The answer should be x-5. If you plug in 5 however, the denominator becomes 5-5=0 which is undefined since you cannot divide by 0.

• i wrote the question wrong ,sorry about that – steve May 23 at 1:33