Question

Phân tích đa thức thành nhân tử:

a. \(2x^3-x^2+5x+3\)

b. \(x^3+5x^2+8x+4\)

c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

d. \(4x^4+1\)

e. \(x^4-7x^3+14x^2-7x+1\)

Answer 1

mk ghi đáp án, còn lại bạn tự biến đổi

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

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

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

d) \(4x^4+1=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)

e) \(x^4-7x^3+14x^2-7x+1=\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)

Answer 2

mk làm chi tiết theo yêu của của người hỏi đề:

a) \(2x^3-x^2+5x+3\)

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

\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)

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

b)  \(x^3+5x^2+8x+4\)

\(=\left(x^3+4x^2+4x\right)+\left(x^2+4x+4\right)\)

\(=x\left(x^2+4x+4\right)+\left(x^2+4x+4\right)\)

\(=\left(x+1\right)\left(x^2+4x+4\right)\)

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

Answer 3

d)  \(4x^4+1=4x^4+4x^2+1-4x^2\)

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

e) \(x^4-7x^3+14x^2-7x+1\)

\(=\left(x^4-3x^3+x^2\right)-\left(4x^3-12x^2+4x\right)+\left(x^2-3x+1\right)\)

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

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

Answer 4

c)  \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt     \(x^2+7x+11=t\)

Khi đó ta có:

\(A=\left(t-1\right)\left(t+1\right)-24=t^2-25=\left(t+5\right)\left(t-5\right)\)

Thay trở lại ta đc:

\(A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

     \(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)