Question

[Đỗ Thu Hoài]

Phân tích đa thức thành nhân tử

a)\(a^3+3a^2+3a+1-27b\)

b)\(64x^3y+\frac{1}{4}yz^3\)

c)\(x^6-x^4+2x^3+2x^2\)

d)\(2x^3-x^2-1\)

Các bạn giúp mik nhé mik đang cần gấp

Answer 1

[lý canh hy]

c, \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

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

Answer 2

[lý canh hy]

d,

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

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

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

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