Question

phân tích đa thức thành nhân tử A=8abc+4(ab+bc+ac)+2(a+b+c)+1

Answer 1

Ta có: 

\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)

\(A=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)

\(A=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)

\(A=\left(2c+1\right)\left(4ab+2a+2b+1\right)\)

\(A=\left(2c+1\right)\left[2a\left(2b+1\right)+\left(2b+1\right)\right]\)

\(A=\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)\)

Answer 2

Ta có:\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)

\(=8abc+4ab+4bc+4ca+2a+2b+2c+1\)

\(=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)

\(=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)

\(=\left(2c+1\right)\left(4ab+2b+2a+1\right)\)

\(=\left(2c+1\right)\left[2b\left(2a+1\right)+\left(2a+1\right)\right]\)

\(=\left(2c+1\right)\left(2b+1\right)\left(2a+1\right)\)