Bài 1:
\(a,4x+12x^2=4x\left(1+3x\right)\\
b,x^3-8+3x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4\right)+3x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4+3x\right)=\left(x-2\right)\left(x^2+5x+4\right)=\left(x-2\right)\left[\left(x^2+x\right)+\left(4x+4\right)\right]=\left(x-2\right)\left[x\left(x+1\right)+4\left(x+1\right)\right]=\left(x-2\right)\left(x+1\right)\left(x+4\right)\\
c,x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x-2y+1\right)\left(x+2y+1\right)\)
Bài 2:
\(a,x^2+5x-3x-15=0\\
\Rightarrow x\left(x+5\right)-3\left(x+5\right)=0\\
\Rightarrow\left(x-3\right)\left(x+5\right)=0\\
\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
