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

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

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