Question

[Lê Quốc Anh]

Phân tích đa thức thành nhân tử:

a) 2x³ - x² + 5x +3

b) 27x³ - 27x² + 18x - 4

c) 4x⁴ - 32x² + 1

d) 3(x⁴ + x² + 1) - (x² + x + 1)²

e) 4x⁴ + 4x³ + 5x² + 2x + 1

f) x⁸ + x + 1

g) 3x² + 22xy + 11x + 37y + 7y² + 10

Answer 1

[Không Tên]

\(x^8+x+1\)

\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

Answer 2

[Đặng Khánh Linh]

a)\(2x^3-x^2+5x+3=2x^3-2x^2+x^2+6x-x+3\)

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

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

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

b)\(27x^3-27x^2+18x-4=27x^3-18x^2-9x^2+12x+6x-4\)

\(=3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\)

\(=\left(9x^2-6x+4\right)\left(3x-1\right)\)

c)\(4x^4-32x^2+1=4x^4+4x^2-36x^2+1\)

\(=\left(4x^4+4x^2+1\right)-36x^2\)

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

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