Question

Phân tích đa thức thành nhân tử

a) (x+1)(x+3)(x+5)(x+7)+15

b) \(x^3+4x^2-5x\)

c) \(x^3-5x^2+8x-4\)

Answer 1

\(a.\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left(x+1\right)\left(x+7\right).\left(x+3\right)\left(x+5\right)+15\)

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

\(=\left(x^2+8x+11-4\right)\left(x^2+8x+11+4\right)+15\)

\(=\left(x^2+8x+11\right)^2-4^2+15\)

\(=\left(x^2+8x+11\right)-1\)

\(=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+2x+6x+12\right)\)

\(=\left(x^2+8x+10\right)\left[x\left(x+2\right)+6\left(x+2\right)\right]\)

\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

\(b.x^3+4x^2-5x=x^3-x^2+5x^2-5x=x^2\left(x-1\right)+5x\left(x-1\right)=\left(x-1\right)\left(x^2+5x\right)=x\left(x-1\right)\left(x+5\right)\)

\(c.x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)

Answer 2

a) Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=\left(\left(x+1\right)\left(x+7\right)\right)\cdot\left(\left(x+3\right)\left(x+5\right)\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(t=x^2+8x\), A trở thành:

\(\left(t+7\right)\left(t+15\right)+15=t^2+22t+120=\left(t+10\right)\left(t+12\right)\)

\(\Rightarrow A=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x+4-\sqrt{6}\right)\left(x+4+\sqrt{6}\right)\left(x+2\right)\left(x+6\right)\)

Kl: \(A=\left(x+4-\sqrt{6}\right)\left(x+4+\sqrt{6}\right)\left(x+2\right)\left(x+6\right)\)

b) \(x^3+4x^2-5x=x\left(x^2+4x-5\right)=x\left(x-1\right)\left(x+5\right)\)

c) \(x^3-5x^2+8x-4=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)