Question

Phân tích đa thức thành nhân tử:

\(x^2+12x+36=0\)

\(4x^2-4x+1=0\)

\(x^3+6x^2+12x+8=0\)

Answer 1

a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

=>\(\left(2x-1\right)^2=0\)

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

=>\(\left(x+2\right)^3=0\)

=>x+2=0

=>x=-2