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

1 Answer
Apr 25, 2018

Answer:

#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)#