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

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

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Answer 1 · 2016-05-21T10:10:21

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

Explanation:

We can write $12 j^{2} k = 2^{2} \cdot 3 \cdot j^{2} \cdot k$ $12j^2k=2^2*3*j^2*k$

$36 \cdot j^{6} \cdot k^{6} = 2^{2} \cdot 3^{2} \cdot j^{6} \cdot k^{6}$ $36*j^6*k^6=2^2*3^2*j^6*k^6$ and

$12 j^{2} = 2^{2} \cdot 3 \cdot j^{2}$ $12j^2=2^2*3*j^2$

Hence $12 j^{2} k - 36 \cdot j^{6} \cdot k^{6} + 12 j^{2}$ $12j^2k-36*j^6*k^6+12j^2$

= $2^{2} \cdot 3 \cdot j^{2} \cdot k - 2^{2} \cdot 3^{2} \cdot j^{6} \cdot k^{6} + 2^{2} \cdot 3 \cdot j^{2}$ $2^2*3*j^2*k-2^2*3^2*j^6*k^6+2^2*3*j^2$

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

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

= $2^{2} \cdot 3 \cdot j^{2} (k - 3 j^{4} \cdot k^{6} + 1)$ $2^2*3*j^2(k-3j^4*k^6+1)$

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

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

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