Question

Answer 1

\(4x^4+1\)

\(=4x^4+4x^2+1-4x^2\)

\(=\left(4x^4+4x^2+1\right)-4x^2\)

\(=\left(2x^2+1\right)^2-4x^2\)

\(=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)

\(16x^4+4\)

\(=4\left(4x^4+1\right)\)

\(=4\left(4x^4+4x^2+1-4x^2\right)\)

\(=4\left[\left(4x^4+4x^2+1\right)-\left(2x\right)^2\right]\)

\(=4\left[\left(2x^2+1\right)^2-\left(2x\right)^2\right]\)

\(=4\left(2x^2+1-2x\right)\left(2x^2+1+2x\right)\)

Answer 2

f) $4x^4+1$

$=(4x^4+4x^2+1)-4x^2$

$=[(2x^2)^2+2\cdot2x^2\cdot1+1^2]-(2x)^2$

$=(2x^2+1)^2-(2x)^2$

$=(2x^2+1-2x)(2x^2+1+2x)$

g) $16x^4+4$

$=(16x^4+16x^2+4)-16x^2$

$=[(4x^2)^2+2\cdot4x^2\cdot2+2^2]-(4x)^2$

$=(4x^2+2)^2-(4x)^2$

$=(4x^2+2-4x)(4x^2+2+4x)$

$=[2(2x^2-2x+1)][2(2x^2+2x+1)]$

$=4(2x^2-2x+1)(2x^2+2x+1)$