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How do you simplify #( t^2-49)/(t^2+4t-21) * (t^2+8t+15)/(t^2-2t-35)#?

Answer 1

Eli Assouline

You first factorise, and then see what you can cancel.
Use 'special products' if needed.

#=((t+7)(t-7))/((t+7)(t-3))*((t+3)(t+5))/((t-7)(t+5))#

Now cancel:

#=(cancel((t+7))cancel((t-7)))/(cancel((t+7))(t-3))*((t+3)cancel((t+5)))/(cancel((t-7))cancel((t+5)))#

#=(t+3)/(t-3)#

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Answer 2

Eli Assouline

To simplify the expression ( t^2-49)/(t^2+4t-21) * (t^2+8t+15)/(t^2-2t-35), we can factorize the numerator and denominator of each fraction and then cancel out any common factors.

The numerator of the first fraction, t^2-49, can be factored as (t+7)(t-7), and the denominator, t^2+4t-21, can be factored as (t+7)(t-3).

The numerator of the second fraction, t^2+8t+15, can be factored as (t+3)(t+5), and the denominator, t^2-2t-35, can be factored as (t-7)(t+5).

Now, we can cancel out the common factors in the numerator and denominator:

(t+7)(t-7)/(t+7)(t-3) * (t+3)(t+5)/(t-7)(t+5)

After canceling out the common factors, we are left with:

(t-7)/(t-3) * (t+3)

Therefore, the simplified expression is (t-7)/(t-3) * (t+3).

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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.