Question

How do you factor `32z^2 - 2t^4z^2`?

Answer 1

`2z^2(2-t)(2+t)(t+2i)(t-2i)`

Explanation:

`"take out a "color(blue)"common factor "2z^2`

`=2z^2(16-t^4)`

`16-t^4" is a "color(blue)"difference of squares"`

`•color(white)(x)a^2-b^2=(a-b)(a+b)`

`"here "a=4" and "b=t^2`

`16-t^4=(4-t^2)(4+t^2)`

`4-t^2" is also a "color(blue)"difference of squares"`

`"here "a=2" and "b=t`

`rArr4-t^2=(2-t)(2+t)`

`"we can factor "4+t^2" by solving "4+t^2=0`

`4+t^2=0rArrt^2=-4rArrt=+-2i`

`rArr4+t^2=(t-2i)(t+2i)`

`rArr32z^2-2t^4z^2=2z^2(2-t)(2+t)(t+2i)(t-2i)`