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How do you simplify #(x^2+4x-32)/(x+5) * (x-3)/(x^2-7x+12)#?

Answer 1

Abigail Allen

You first factorise the quatratic parts, and then you may cancel some terms

#=((x+8)(x-4))/(x+5)*(x-3)/((x-3)(x-4))#

#=((x+8)cancel((x-4))cancel((x-3)))/((x+5)cancel((x-3))cancel((x-4)))#

#=(x+8)/(x+5)#

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Answer 2

Eva Adamson

To simplify the expression (x^2+4x-32)/(x+5) * (x-3)/(x^2-7x+12), we can first factor the numerator and denominator of each fraction.

The numerator of the first fraction, x^2+4x-32, can be factored as (x+8)(x-4). The denominator of the first fraction, x+5, cannot be factored further.

The numerator of the second fraction, x-3, cannot be factored further. The denominator of the second fraction, x^2-7x+12, can be factored as (x-3)(x-4).

Now, we can cancel out common factors between the numerators and denominators.

(x+8)(x-4)/(x+5) * (x-3)/(x-3)(x-4)

After canceling out the common factor (x-4) and (x-3), we are left with:

(x+8)/(x+5)

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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.