Mathematics OpenStudy (anonymous):

Factor: 6(x^2 - 4x + 4)^2 + (x^2 - 4x + 4) - 1 OpenStudy (netlopes1):

first: $(x^2-4x+4)=(x-2)^2$ OpenStudy (anonymous):

Okay OpenStudy (netlopes1):

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? OpenStudy (anonymous):

If (x-2)^2 = k then wouldn't it just be 6k + k - 1 OpenStudy (netlopes1):

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? OpenStudy (netlopes1):

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

I think I see OpenStudy (pfenn1):

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

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: $6.[(x-2)^2-1/3].[(x-2)^2+1/2]$

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