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\({b^2 + 2b -24 \over 2b^2 - 72}\) and show each step :)

Guest Mar 15, 2018

Best Answer 

 #1
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Simplify the following:

(b^2 + 2 b - 24)/(2 b^2 - 72)

 

The factors of -24 that sum to 2 are 6 and -4. So, b^2 + 2 b - 24 = (b + 6) (b - 4):

((b + 6) (b - 4))/(2 b^2 - 72)

 

Factor 2 out of 2 b^2 - 72:

((b + 6) (b - 4))/(2 (b^2 - 36))

 

b^2 - 36 = b^2 - 6^2:

((b + 6) (b - 4))/(2 (b^2 - 6^2))

 

Factor the difference of two squares. b^2 - 6^2 = (b - 6) (b + 6):

((b + 6) (b - 4))/(2 (b - 6) (b + 6))

 

((b + 6) (b - 4))/(2 (b - 6) (b + 6)) = (b + 6)/(b + 6)×(b - 4)/(2 (b - 6)) = (b - 4)/(2 (b - 6)):

 

(b - 4) / (2(b - 6))

Guest Mar 15, 2018
 #1
avatar
0
Best Answer

Simplify the following:

(b^2 + 2 b - 24)/(2 b^2 - 72)

 

The factors of -24 that sum to 2 are 6 and -4. So, b^2 + 2 b - 24 = (b + 6) (b - 4):

((b + 6) (b - 4))/(2 b^2 - 72)

 

Factor 2 out of 2 b^2 - 72:

((b + 6) (b - 4))/(2 (b^2 - 36))

 

b^2 - 36 = b^2 - 6^2:

((b + 6) (b - 4))/(2 (b^2 - 6^2))

 

Factor the difference of two squares. b^2 - 6^2 = (b - 6) (b + 6):

((b + 6) (b - 4))/(2 (b - 6) (b + 6))

 

((b + 6) (b - 4))/(2 (b - 6) (b + 6)) = (b + 6)/(b + 6)×(b - 4)/(2 (b - 6)) = (b - 4)/(2 (b - 6)):

 

(b - 4) / (2(b - 6))

Guest Mar 15, 2018

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