0
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ 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.” $\endgroup$ – gen-z ready to perish May 23 at 2:14
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ i wrote the question wrong ,sorry about that $\endgroup$ – steve May 23 at 1:33
  • 1
    $\begingroup$ 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. $\endgroup$ – Ross Millikan May 23 at 1:39
  • $\begingroup$ Edited to reflect updated question. $\endgroup$ – vanPelt2 May 23 at 2:45
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ i wrote the question wrong ,sorry about that $\endgroup$ – steve May 23 at 1:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.