Bài 2:
i: \(x^2+8x+15=\left(x+3\right)\left(x+5\right)\)
j: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
\(a,=2x\left(y^3-3x+5y\right)\\ b,=\left(x-1\right)^3-y^3=\left(x-y-1\right)\left(x^2-2x+1+xy+y+y^2\right)\\ c,=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\\ d,=a^5\left(a-1\right)-2a^2\left(a-1\right)=a^2\left(a^3-2\right)\left(a-1\right)\\ e=\left(5-x^2\right)\left(25+5x^2+x^4\right)\\ f,=4\left(x^4+x^2y^2-2y^4\right)\\ =4\left(x^4-x^2y^2+2x^2y^2-2y^4\right)\\ =4\left[x^2\left(x^2-y^2\right)+2y^2\left(x^2-y^2\right)\right]\\ =4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\\ g,=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)
\(h,=\left(x-1\right)\left(x^2-16\right)=\left(x-4\right)\left(x-1\right)\left(x+4\right)\\ i,=x^2+3x+5x+15=\left(x+3\right)\left(x+5\right)\\ j,=x^2+3x-4x-12=\left(x+3\right)\left(x-4\right)\\ k,=x^2+x+3x+3=\left(x+1\right)\left(x+3\right)\\ l,=4x^2-2x+6x-3=2x\left(2x-1\right)+3\left(2x-1\right)\\ =\left(2x+3\right)\left(2x-1\right)\)
