Question

Phân tích các đa thức sau thành nhân tử

a, x³ - x² - x + 1

b, x³ + x² - 4x - 4

c, a⁵ + 27a²

d, x⁴ - 8x

e, x⁴ - 4x³ + 4x²

f, 2x⁴ - 32

Answer 1

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

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

c)\(a^5+27a^2=a^2\left(a^3+27\right)=a^2\left(a+3\right)\left(a^2-3a+9\right)\)

d)\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)

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

f)\(2x^4-32=2\left(x^4-16\right)=2\left(x^2+4\right)\left(x^2-4\right)=2\left(x^2+4\right)\left(x+2\right)\left(x-2\right)\)

Answer 2

a) \(x^3-x^2-x+1\)

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

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

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

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

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

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

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

c) \(a^5+27a^2=a^2\left(a^3+27\right)\)

\(=a^2\left(a+3\right)\left(a^2-3a+9\right)\)

d) \(x^4-8x=x\left(x^3-8\right)\)

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

e) \(x^4-4x^3+4x^2\)

\(=\left(x^2\right)^2-2\cdot x^2\cdot2x+\left(2x\right)^2\)

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

f) \(2x^4-32=2\left(x^4-16\right)\)

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

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