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(2ax + bx)2
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You'd have to distribute the 2 to your whole equation. 2ax*2=4ax bx*2=2bx So now you have 4ax+2bx.
\[(2ax+bx)^2=(2ax+bx)(2ax+bx)\] Compare to \[(u+v)^2=u^2+2uv+v^2\]
I'm assuming the 2 at the end was an exponent and not multiplication.
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