Factor: 6(x^2 - 4x + 4)^2 + (x^2 - 4x + 4) - 1

first: \[(x^2-4x+4)=(x-2)^2\]

Okay

Now call \[(x-2)^2=k\] or other any letter . Look the "new" expression: \[6k^2+k-1 \rightarrow (k-1/3).(k+1/2)\] or the two factors \[[(x-2)^2-1/3].[(x-2)^2+1/2)\] ok?

If (x-2)^2 = k then wouldn't it just be 6k + k - 1

Please note that (x^2-4x+4) only this is same (x-2)^2 and this factor is still elevated to 2. Look the original question: 6(x^2 - 4x + 4)^2 .Did you see the detail?

If you called (x^2-4x+4) OR (x-2)^2 like "k", we have 6K^2 +k-1.

I think I see

I agree with @netlopes1, except this step:T\[6k^2+k-1= (2k+1)(3k-1)\]And when you substitute for k you get\[6(x^2 - 4x + 4)^2 + (x^2 - 4x + 4) - 1=[2(x-2)^2+1][3(x-2)^2-1]\]

Later is better than never!! OK, @pfenn1, you are correct. Congratulations!! My answer would be correct if i've included the numer "6", like this: \[6.[(x-2)^2-1/3].[(x-2)^2+1/2]\]