# factorizing

• Oct 31st 2008, 11:33 PM
jvignacio
factorizing
$\displaystyle (2x+1)^2 + x(2+4x) = 0$

then i expanded and collected like terms and got

$\displaystyle 8x^2+6x+1 = 0$

then i was having problems factorizing this expression.any help?
• Nov 1st 2008, 12:17 AM
Peritus
you don't need to expand anything:

$\displaystyle \begin{gathered} \left( {2x + 1} \right)^2 + x\left( {2 + 4x} \right) = 0 \hfill \\ \Leftrightarrow \left( {2x + 1} \right)^2 + 2x\left( {2x + 1} \right) = 0 \hfill \\ \Leftrightarrow \left( {2x + 1} \right)\left( {4x + 1} \right) = 0 \hfill \\ \end{gathered}$
• Nov 1st 2008, 12:24 AM
jvignacio
Quote:

Originally Posted by Peritus
you don't need to expand anything:

$\displaystyle \begin{gathered} \left( {2x + 1} \right)^2 + x\left( {2 + 4x} \right) = 0 \hfill \\ \Leftrightarrow \left( {2x + 1} \right)^2 + 2x\left( {2x + 1} \right) = 0 \hfill \\ \Leftrightarrow \left( {2x + 1} \right)\left( {4x + 1} \right) = 0 \hfill \\ \end{gathered}$

yes yes i had it right, just made a silly mistake... thanks for that!