$\displaystyle

16x^3+16x^2+3x

$

$\displaystyle

x(16x^2+16x+3)

$

What else can I do?

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- Nov 11th 2008, 05:50 AM #1

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- Nov 11th 2008, 05:55 AM #2

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- Nov 11th 2008, 07:07 AM #3

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- Nov 11th 2008, 07:27 AM #4

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No, Tweety.

You must have made a mistake, since it is $\displaystyle 16x^3$, not $\displaystyle 16x^2$. But anyway you helped me to find the right answer. Thanks.

$\displaystyle

16x^3+16x^2+3x

$

$\displaystyle

16x^3+12x^2+4x^2+3x

$

$\displaystyle

4x^2(4x+3)+x(4x+3)

$

$\displaystyle

(4x^2+x)(4x+3)

$