Question

Phân tích đa thức thành nhân tử

a,x^2+3x+2

b,x^4+4

c,x^2y^2+1-x^2-y^2

Answer 1

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

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

Answer 2

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