Question

Phân tích đa thức thành nhân tử :

x⁸ + 3x⁴ + 4

3x² + 22xy + 11x + 37y + 7y² + 10

Answer 1

\(x^8+3x^4+4\)

\(=x^8+4x^4+4-x^4\)

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

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

\(3x^2+22xy+11x+37y+7y^2+10\)

\(=\left(3x^2+6x+21xy\right)+\left(7y^2+xy+2y\right)+\left(5x+35y+10\right)\)

\(=3x\left(x+7y+2\right)+y\left(x+7y+2\right)+5\left(x+7y+2\right)\)

\(=\left(3x+y+5\right)\left(x+7y+2\right)\)

Answer 2

\(x^8+3x^4+4\)

\(=x^8+4x^4+4-x^4\)

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

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