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How do you divide #(s^2-3s)/(s^2-s-6) div (s-6)/(s+2)#?

Answer 1

Eva Abramson

#= ((s^2-3s)/(s^2-s-6))/((s-6)/(s+2)) #

#= ((s^2-3s)(s+2))/((s^2-s-6)(s-6)) #

#= ((s^2-3s)(s+2))/((s-3)(s+2)(s-6))#

#= ((s^2-3s)cancel((s+2)))/((s-3)cancel((s+2))(s-6))#

#= ((s^2-3s))/((s-3)(s-6))#

#= ((s^2-3s))/((s^2-9s+18))#

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Answer 2

Eva Adamson

To divide the expression (s^2-3s)/(s^2-s-6) by (s-6)/(s+2), we can multiply the first expression by the reciprocal of the second expression.

First, factor the denominators: (s^2-s-6) can be factored as (s-3)(s+2), and (s-6) can be factored as (s-6).

Next, rewrite the division as multiplication by the reciprocal: (s^2-3s)/(s^2-s-6) * (s+2)/(s-6).

Now, cancel out the common factors: (s+2) in the numerator and (s-6) in the denominator.

The simplified expression is (s-3)/(s-3), which equals 1.

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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.