Ta có: \(A=\left(x^2+4x\right)^2-2\left(x^2+4x\right)-15\)
Đặt \(x^2+4x=y\) , Ta được:
\(A=y^2-2y-15=y^2-5y+3y-15\)
\(=\left(y^2+3y\right)-\left(5y+15\right)\)
\(=y\left(y+3\right)-5\left(y+3\right)\)
\(=\left(y-5\right).\left(y+3\right)\)
Thay \(x^2+4x=y\) , Ta được:
\(A=\left(x^2+4x-5\right).\left(x^2+4x+3\right)\)
\(\Rightarrow A=\left(x^2+x+3x+3\right).\left(x^2+5x-x-5\right)\)
\(\Rightarrow A=\left[\left(x^2+x\right)+\left(3x+3\right)\right].\left[\left(x^2-x\right)+\left(5x-5\right)\right]\)
\(\Rightarrow A=\left[x\left(x+1\right)+3\left(x+1\right)\right].\left[x\left(x-1\right)+5\left(x-1\right)\right]\)
\(\Rightarrow A=\left[\left(x+1\right)\left(x+3\right)\right].\left[\left(x-1\right)\left(x+5\right)\right]\)
\(\Rightarrow A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x-1\right)\)
Chúc bạn học tốt !!!
\(A=\left(x^2+4x\right)^2-2\left(x^2+4x\right)-15\)
\(=\left[\left(x^2+4x\right)^2-2\left(x^2+4x\right)+1\right]-16\)
\(=\left(x^2+4x-1\right)^2-4^2\)
\(=\left(x^2+4x-1+4\right)\left(x^2+4x-1-4\right)\)
\(=\left(x^2+4x+3\right)\left(x^2+4x-5\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x-1\right)\left(x+5\right)\)
