How do you factor #196- n ^ { 6} p ^ { 4} q ^ { 2}#?

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Answer 1 · 2017-06-19T15:40:07

See a solution process below:

This is a special form of the quadratic:

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

Substitute:

#14# for #a#

#n^3p^2q# for #b#

Giving:

#(14 + n^3p^2q)(14 - n^3p^2q) = 14^2 - (n^3p^2q)^2#

#(14 + n^3p^2q)(14 - n^3p^2q) = 196 - n^6p^4q^2#

Answer 2 · 2017-06-19T15:53:29

#(14 + n^3p^2q) xx (14 - n^3p^2q)#

#196 - n^6 p^4 q^2#

Solution

Look for common factors to factorize

#^2# is common

#:.# #(14^2 + n^(2xx3) xx p^(2xx2) xx q^2)#

#[(14xx14) + (n^3xxn^3) xx (p^2xxp^2) xx (qxxq)]#

Factorize

#(14 + n^3p^2q) xx (14 - n^3p^2q)#