How do you factor #12j^2k - 36j^6k^6 + 12j^2#?

1 Answer
May 21, 2016

Answer:

#12j^2k-36j^6k^6+12j^2=12j^2(k-3j^4k^6+1)#

Explanation:

We can write #12j^2k=2^2*3*j^2*k#

#36*j^6*k^6=2^2*3^2*j^6*k^6# and

#12j^2=2^2*3*j^2#

Hence #12j^2k-36*j^6*k^6+12j^2#

= #2^2*3*j^2*k-2^2*3^2*j^6*k^6+2^2*3*j^2#

Now minimum power for #2# is #2#; for #3# is #1#; for #j# is #2# and for #k# is not there in last moomial. Taking this as common, we get

#12j^2k-36j^6k^6+12j^2#

= #2^2*3*j^2(k-3j^4*k^6+1)#

= #12j^2(k-3j^4k^6+1)#