Question

 Phân tích đa thức thành nhân tử : (x² + 6x – 5)(x² + 6x + 3) – 20

Answer 1

Ta có: (x²+6x-5)(x²+6x+3)-20

      = [(x²+6x-1)-4][(x²+6x-1)+4]-20

      = (x²+6x-1)²-16-20

      = (x²+6x-1)²-36

      = (x²+6x-7)(x²+6x-5)

      = (x+7)(x-1)(x²+6x-5)

Answer 2

\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)\\ =\left(x^2+6x-1\right)^2-16-20\\ =\left(x^2+6x-1\right)^2-36\\ =\left(x^2+6x-1-6\right)\left(x^2+6x-1+6\right)\\ =\left(x^2+6x-7\right)\left(x^2+6x+5\right)\\ =\left(x-1\right)\left(x+7\right)\left(x+1\right)\left(x+5\right)\)

Answer 3

\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)-20\)

\(=\left(x^2+6x\right)^2-2\left(x^2+6x\right)-35\)

\(=\left(x^2+6x-7\right)\left(x^2+6x+5\right)\)

\(=\left(x+7\right)\left(x-1\right)\left(x+1\right)\left(x+5\right)\)