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

Jun 19, 2017

See a solution process below:

Explanation:

This is a special form of the quadratic:

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

Substitute:

$14$ for $a$

${n}^{3} {p}^{2} q$ for $b$

Giving:

$\left(14 + {n}^{3} {p}^{2} q\right) \left(14 - {n}^{3} {p}^{2} q\right) = {14}^{2} - {\left({n}^{3} {p}^{2} q\right)}^{2}$

$\left(14 + {n}^{3} {p}^{2} q\right) \left(14 - {n}^{3} {p}^{2} q\right) = 196 - {n}^{6} {p}^{4} {q}^{2}$

Jun 19, 2017

$\left(14 + {n}^{3} {p}^{2} q\right) \times \left(14 - {n}^{3} {p}^{2} q\right)$

Explanation:

$196 - {n}^{6} {p}^{4} {q}^{2}$

Solution

Look for common factors to factorize

$^ 2$ is common

$\therefore$ $\left({14}^{2} + {n}^{2 \times 3} \times {p}^{2 \times 2} \times {q}^{2}\right)$

$\left[\left(14 \times 14\right) + \left({n}^{3} \times {n}^{3}\right) \times \left({p}^{2} \times {p}^{2}\right) \times \left(q \times q\right)\right]$

Factorize

$\left(14 + {n}^{3} {p}^{2} q\right) \times \left(14 - {n}^{3} {p}^{2} q\right)$