Question

Phân tích đa thức thành nhân tử

a, ( x² + 1 )² - 4x²

b, 1 - 4x²

c, 3y² - 3z² + 3x² + 6xy

đ, x⁵ - 3x⁴ + 3x³ - x²

E, 3x² - 6x - 2x +x

Answer 1

a,

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

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

b,

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

c,

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

\(=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)

d,

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

e,

\(3x^2-6x-2x+x=3x^2-7x=x\left(3x-7\right)\)