Question

Phân tích thành nhân tử:

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+12\right)-165x^2\)

Answer 1

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+12\right)-165x^2\)

\(=\left[\left(x+2\right)\left(x+12\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]-165x^2\)

\(=\left(x^2+14x+24\right)\left(x^2+10x+24\right)-165x^2\)

\(=\left(x^2+12x+24+2x\right)\left(x^2+12x+24-2x\right)-165x^2\)

\(=\left(x^2+12x+24\right)^2-4x^2-165x^2\)

\(=\left(x^2+12x+24\right)^2-169x^2\)

\(=\left(x^2+12x+24-13x\right)\left(x^2+12x+24+13x\right)\)

\(=\left(x^2-x+24\right)\left(x^2+25x+24\right)\)

\(=\left(x^2-x+24\right)\left(x^2+x+24x+24\right)\)

\(=\left(x^2-x+24\right)\left[x\left(x+1\right)+24\left(x+1\right)\right]\)

\(=\left(x^2-x+24\right)\left(x+1\right)\left(x+24\right)\)