im terrible at factoring..please help?

(4x+1)^2 - (16x^2 + 4x)

i treated it as a difference of squares

so i did:

(4x+1 + 16x^2+4x)(4x+1 - 16x^2 - 4x)

which i further simplified to:

(16x^2 +8x+1)(-16x^2 +1)

and im pretty sure this isnt right.

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- Jan 13th 2008, 10:10 AMcheckmarksone factoring question
im terrible at factoring..please help?

(4x+1)^2 - (16x^2 + 4x)

i treated it as a difference of squares

so i did:

(4x+1 + 16x^2+4x)(4x+1 - 16x^2 - 4x)

which i further simplified to:

(16x^2 +8x+1)(-16x^2 +1)

and im pretty sure this isnt right. - Jan 13th 2008, 11:33 AMSoroban
Hello, checkmarks!

Quote:

Factor: .$\displaystyle (4x+1)^2 - (16x^2 + 4x)$

i treated it as a difference of squares . . . . $\displaystyle {\color{blue}(16x^2 + 4x)}$ is**not**a square

Take out the common factor: .$\displaystyle (4x+1)\bigg[(4x+1) - 4x\bigg] $

And we have: .$\displaystyle (4x + 1)[4x + 1 - 4x) \;\;=\;\;(4x+1)(1) \;\;=\;\;\boxed{4x + 1}$