\(\text{a) }x^2+4x+3\\ =x^2+x+3x+3\\ =\left(x^2+x\right)+\left(3x+3\right)\\ =x\left(x+1\right)+3\left(x+1\right)\\ =\left(x+3\right)\left(x+1\right)\\ \)
\(\text{b) }x^2-13x+12\\ =x^2-x-12x+12\\ =\left(x^2-x\right)-\left(12x-12\right)\\ =x\left(x-1\right)-12\left(x-1\right)\\ =\left(x-12\right)\left(x-1\right)\\ \)
\(\text{c) }x^2+5x-6\\ =x^2-x+6x-6\\ =\left(x^2-x\right)+\left(6x-6\right)\\ =x\left(x-1\right)+6\left(x-1\right)\\ =\left(x-1\right)\left(x+1\right)\\ \)
\(\text{d) }2x^2+3x-5\\ =2x^2-2x+5x-5\\ =\left(2x^2-2x\right)+\left(5x-5\right)\\ =2x\left(x-1\right)+5\left(x-1\right)\\ =\left(2x+5\right)\left(x-1\right)\\ \)
\(\text{e) }a^{m+3}-a^m+a-1\\ =\left(a^{m+3}-a^m\right)+\left(a-1\right)\\ =a^m\left(a^3-1\right)+\left(a-1\right)\\ =a^m\left(a-1\right)\left(a^2+a+1\right)+\left(a-1\right)\\ =\left(a-1\right)\left[a^m\left(a^2+a+1\right)+1\right]\\ =\left(a-1\right)\left(a^{m+2}+a^{m+1}+1\right)\\ \)
\(x^2+4x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+3\right)\left(x+1\right)\)
\(x^2-13x+12=x\left(x-12\right)-\left(x-12\right)=\left(x-12\right)\left(x-1\right)\)
\(x^2+3x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x-2\right)\left(x+5\right)\)
\(x^2+5x-6=x\left(x-6\right)+\left(x-6\right)=\left(x-6\right)\left(x+1\right)\)
\(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
\(a^{m+3}-a^m+a-1=a^m\left(a^3-1\right)+\left(a-1\right)=a^m\left(a-1\right)\left(a^2+a+1\right)+\left(a-1\right)=\left(a-1\right)\left[a^m\left(a^2+a+1\right)+1\right]=\left(a-1\right)\left(a^{m+2}+a^{m+1}+a^m+1\right)\)
