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

Answer 1

#u+9#

#u != 2#

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#

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Answer 2

Nathan Axel

To simplify the expression (u^2+7u-18)/(u-2), we can factor the numerator and then cancel out any common factors with the denominator. The factored form of the numerator is (u+9)(u-2). Therefore, the simplified expression is (u+9).

The excluded value for the variable u is 2, as it would result in division by zero.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.