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How do you simplify #[(x^2+2x)(x^2+2x-3)]/[(x^2+x-2)(x^2+3x)]#?

Answer 1

Hazel Anglin

Since

#x^2+2x=x(x+2)# #x^2+2x-3=(x+3)(x-1)# #x^2+x-2=(x+2)(x-1)# #x^2+3x=x(x+3)#

you would substitute in the given fraction and symplify:

#((x^2+2x)(x^2+2x-3))/((x^2+x-2)(x^2+3x))=(x(x+2)(x+3)(x-1))/((x+2)(x-1)x(x+3))=1#

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Answer 2

Jace Anderson

To simplify the expression [(x^2+2x)(x^2+2x-3)]/[(x^2+x-2)(x^2+3x)], we can factorize the numerator and denominator.

The numerator can be factored as (x(x+2))(x^2+2x-3).

The denominator can be factored as (x-1)(x+2)(x(x+3)).

Now, we can cancel out common factors between the numerator and denominator.

Canceling out (x(x+2)) and (x+2) from the numerator and denominator respectively, we get:

(x(x+2)(x^2+2x-3))/((x-1)(x(x+3)))

Further simplifying the expression, we have:

(x(x+2)(x-1)(x+3))/(x-1)(x(x+3))

Finally, we can cancel out the common factor (x-1) from the numerator and denominator, resulting in:

(x(x+2)(x+3))/x(x+3)

Simplifying further, we are left with:

(x+2)

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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.