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How do you simplify #(2x^2-56)/(x^2-16) - (x+7)/(x+4)#?

Answer 1

Abigail Allen

#(x-7)/(x-4)#

#(2x^2-56)/(x^2-16)-(x+7)/(x+4)#

=#[(2x^2-56)-(x+7)*(x-4)]/(x^2-16)#

=#[(2x^2-56)-(x^2+3x-28)]/(x^2-16)#

=#(x^2-3x-28)/(x^2-16)#

=#[(x-7)(x+4)]/[(x-4)(x+4)]#

=#(x-7)/(x-4)#

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Answer 2

Hazel Anglin

To simplify the expression (2x^2-56)/(x^2-16) - (x+7)/(x+4), we first need to factor the denominators. The denominator x^2-16 can be factored as (x+4)(x-4), and the denominator x+4 remains the same.

Next, we find a common denominator for the two fractions, which is (x+4)(x-4).

Now, we can rewrite the expression with the common denominator:

[(2x^2-56)/(x^2-16)] - [(x+7)/(x+4)] = [(2x^2-56)(x+4) - (x+7)(x-4)] / [(x+4)(x-4)]

Expanding the numerators, we get:

[(2x^3 + 8x^2 - 56x - 224) - (x^2 - 4x + 7x - 28)] / [(x+4)(x-4)]

Combining like terms, we have:

(2x^3 + 8x^2 - 56x - 224 - x^2 + 4x - 7x + 28) / [(x+4)(x-4)]

Simplifying further, we get:

(2x^3 + 7x^2 - 59x - 196) / [(x+4)(x-4)]

Therefore, the simplified expression is (2x^3 + 7x^2 - 59x - 196) / [(x+4)(x-4)].

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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.