Phân tích đa thức (x+1) (x+2) (x+3) (x+4)-24 thành nhân tử - Xuan Xuan

  • 3,000
  • Tác giả: admin
  • Ngày đăng:
  • Lượt xem: 3
  • Tình trạng: Còn hàng

\(f,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(t=x^2+5x+4\) , ta có
\(t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t^2+2t+1\right)-25\)

\(=\left(t+1\right)^2-5^2\)

\(=\left(t+1-5\right)\left(t+1+5\right)\)

\(=\left(t-4\right)\left(t+6\right)\)

\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(g,\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(t=x^2-8x+7\), ta có:

\(t\left(t+8\right)-20\)

\(=t^2+8t-20\)

\(=\left(t^2+8t+16\right)-36\)

\(=\left(t+4\right)^2-6^2\)

\(=\left(t+4+6\right)\left(t+4-6\right)\)

\(=\left(t+10\right)\left(t-2\right)\)

\(=\left(x^2-8x+7+10\right)\left(x^2-8x+7-2\right)\)

\(=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)