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

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Answer 1 · 2016-05-21T10:10:21

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

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

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

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

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

How do you factor 12j^2k - 36j^6k^6 + 12j^212j^2k - 36j^6k^6 + 12j^2?