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I keep getting this wrong!

Guest Jul 8, 2017

Best Answer 

 #1
avatar+7324 
+2

\(\frac{m-4}{m^2+8m+16}-\frac{m-4}{m+4}\)

 

We need to get a common denominator. To do that...multiply the second fraction by \( \frac{m+4}{m+4} \)  .

 

\(=\frac{m-4}{m^2+8m+16}-\frac{(m-4)(m+4)}{(m+4)(m+4)}\)

 

Multiply the numerator and denominator out.

 

\(=\frac{m-4}{m^2+8m+16}-\frac{m^2-16}{m^2+8m+16}\)

 

Now that we have a common denominator, we can combine the fractions.

 

\(=\frac{m-4-(m^2-16)}{m^2+8m+16} \\~\\ =\frac{m-4-m^2+16}{m^2+8m+16} \\~\\ =\frac{\mathbf{-1}m^2+m+\mathbf{12}}{m^2+\mathbf{8}m+\mathbf{16}}\)

hectictar  Jul 8, 2017
 #1
avatar+7324 
+2
Best Answer

\(\frac{m-4}{m^2+8m+16}-\frac{m-4}{m+4}\)

 

We need to get a common denominator. To do that...multiply the second fraction by \( \frac{m+4}{m+4} \)  .

 

\(=\frac{m-4}{m^2+8m+16}-\frac{(m-4)(m+4)}{(m+4)(m+4)}\)

 

Multiply the numerator and denominator out.

 

\(=\frac{m-4}{m^2+8m+16}-\frac{m^2-16}{m^2+8m+16}\)

 

Now that we have a common denominator, we can combine the fractions.

 

\(=\frac{m-4-(m^2-16)}{m^2+8m+16} \\~\\ =\frac{m-4-m^2+16}{m^2+8m+16} \\~\\ =\frac{\mathbf{-1}m^2+m+\mathbf{12}}{m^2+\mathbf{8}m+\mathbf{16}}\)

hectictar  Jul 8, 2017
 #2
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+1

Thank you

Guest Jul 8, 2017

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