How do you factor #m^ { 2} - 4m + 4- n ^ { 2} + 8n - 16#?

1 Answer
George C.
Nov 23, 2017

#m^2-4m+4-n^2+8n-16 = (m-n+2)(m+n-6)#

Explanation:

The difference of squares identity can be written:

#A^2-B^2=(A-B)(A+B)#

Use this with #A=(m-2)# and #B=n-4# as follows:

#m^2-4m+4-n^2+8n-16 = (m^2-4m+4)-(n^2-8n+16)#

#color(white)(m^2-4m+4-n^2+8n-16) = (m-2)^2-(n-4)^2#

#color(white)(m^2-4m+4-n^2+8n-16) = ((m-2)-(n-4))((m-2)+(n-4))#

#color(white)(m^2-4m+4-n^2+8n-16) = (m-n+2)(m+n-6)#

Alternative method

Given:

#m^2-4m+4-n^2+8n-16#

Set #n=0# or equivalently cover up all of the terms involving #n# to find:

#m^2-4m+4-16 = m^2-4m-12 = (m-6)(m+2)#

Set #m=0# or equivalently cover up all of the terms involving #m# to find:

#4-n^2+8n-16 = -n^2+8n-12#

#color(white)(4-n^2+8n-16) = -(n^2-8n+12)#

#color(white)(4-n^2+8n-16) = -(n-6)(n-2)#

#color(white)(4-n^2+8n-16) = (n-6)(-n+2)#

Then combine the factorisations we have found:

#(m-6)# and #(n-6)# combine to form #(m+n-6)#

#(m+2)# and #(-n+2)# combine to form #(m-n+2)#

So:

#m^2-4m+4-n^2+8n-16 = (m+n-6)(m-n+2)#

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