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How do you simplify #(2z^2 - 11z + 15)/(z^2 - 9)#?

Answer 1

Kennedy Adams

#(2z²-11z+15)/(z²-9)=(2(z-5/2))/(z+3)=(2z-5)/(z+3)#

#(2z²-11z+15)/(z²-9)#

#=(2(z²-11/2z+15/2))/((z-3)(z+3))#

#=(2(z²-5/2z-3z+15/2))/((z-3)(z+3))#

#=(2(z(z-5/2)-3(z-5/2)))/((z-3)(z+3))#

#=(2cancel((z-3))(z-5/2))/(cancel((z-3))(z+3))#

#=(2(z-5/2))/(z+3)=(2z-5)/(z+3)#

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Answer 2

Hazel Anglin

First we need to split the middle term

After splitting... you'll get

#(2z^2-6z-5z+15)/(z^2-3^2)#

This is a law of exponents

#a^2-b^2=(a+b)(a-b)#

Factorize

#(2z(z-3)-5(z-3))/((z-3)(z+3))#

You get

#((2z-5)cancel((z-3)))/(cancel((z-3))(z+3))#

You are left with

#color(red)[(2z-5)/(z+3)]#

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Answer 3

Eli Assouline

To simplify the expression (2z^2 - 11z + 15)/(z^2 - 9), we can factor both the numerator and the denominator. The numerator can be factored as (2z - 3)(z - 5), and the denominator can be factored as (z - 3)(z + 3).

Therefore, the expression simplifies to (2z - 3)(z - 5)/(z - 3)(z + 3).

Next, we can cancel out the common factors of (z - 3) in the numerator and denominator.

The simplified expression is (2z - 3)(z - 5)/(z + 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.