How do you factor 32z^2 - 2t^4z^232z22t4z2?

1 Answer
Apr 25, 2018

2z^2(2-t)(2+t)(t+2i)(t-2i)2z2(2t)(2+t)(t+2i)(t2i)

Explanation:

"take out a "color(blue)"common factor "2z^2take out a common factor 2z2

=2z^2(16-t^4)=2z2(16t4)

16-t^4" is a "color(blue)"difference of squares"16t4 is a difference of squares

•color(white)(x)a^2-b^2=(a-b)(a+b)xa2b2=(ab)(a+b)

"here "a=4" and "b=t^2here a=4 and b=t2

16-t^4=(4-t^2)(4+t^2)16t4=(4t2)(4+t2)

4-t^2" is also a "color(blue)"difference of squares"4t2 is also a difference of squares

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

rArr4-t^2=(2-t)(2+t)4t2=(2t)(2+t)

"we can factor "4+t^2" by solving "4+t^2=0we can factor 4+t2 by solving 4+t2=0

4+t^2=0rArrt^2=-4rArrt=+-2i4+t2=0t2=4t=±2i

rArr4+t^2=(t-2i)(t+2i)4+t2=(t2i)(t+2i)

rArr32z^2-2t^4z^2=2z^2(2-t)(2+t)(t+2i)(t-2i)32z22t4z2=2z2(2t)(2+t)(t+2i)(t2i)