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How do you factor #(x^2-81)/( 2x^2-23x+45)#?

Answer 1

#(x+9)/(2x-5)#

Numerator is a Difference of Squares: #x^2 - 81 = (x-9)(x+9)#

Denominator can be factored. I like using Mustang Method. (Youtube: Mustang Method) #2x^2 - 23x + 45 = (2x - 5)(x - 9)#

#(x^2-81)/(2x^2 - 23x + 45) = ((x-9)(x+9))/((2x-5)(x-9))# Cancel Like Factors from Numerator and Denominator # = (x+9)/(2x-5)#

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Answer 2

Genesis Allen

To factor the expression (x^2-81)/(2x^2-23x+45), we can start by factoring the numerator and denominator separately.

The numerator, x^2-81, is a difference of squares and can be factored as (x+9)(x-9).

The denominator, 2x^2-23x+45, cannot be factored further using simple integer factors. We can use the quadratic formula to find the roots of the denominator, which are x = (23 ± √(23^2 - 4(2)(45))) / (2(2)). Simplifying this expression gives us x = (23 ± √(-527)) / 4, which means the denominator does not have any real roots.

Therefore, the expression (x^2-81)/(2x^2-23x+45) cannot be factored any further.

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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.