How do you simplify #(u^2+7u-18)/(u-2)# and what are the excluded values for the variables?

1 Answer
Apr 25, 2017

Answer:

#u+9#

#u != 2#

Explanation:

Although you have a numerator and denominator you are not allowed to simplify yet by cancelling, because there are terms (separated by + and * signs) rather than factors (which are multiplied)

Find the factors first.

#(u^2 +7u-18)/(u-2)#

#((u+9)(u-2))/((u-2))" "larr# now you can cancel

#((u+9)cancel((u-2)))/(cancel((u-2))#

#=u+9#

For excluded values, remember that the denominator may not be equal to #0#.

So, ask yourself what value of #u# would make #0#?

If #u-2=0 " "rarr u =2#