Question

Phân tích đa thức thành nhân tử :

3x³ + 19x² +4x -12

Answer 1

\(3x^3+19x^2+4x-12=3x^3+3x^2+16x^2+16x-12x-12=3x^2\left(x+1\right)+16x\left(x+1\right)-12\left(x+1\right)=\left(x+1\right)\left(3x^2+16x-12\right)=\left(x+1\right)\left(3x^2+18x-2x-12\right)=\left(x+1\right)\left[3x\left(x+6\right)-2\left(x+6\right)\right]=\left(x+1\right)\left(3x-2\right)\left(x+6\right)\)

Answer 2

3x\(^3\)-19x\(^2\)+4x+12

=3x\(^3\)-3x\(^2\)-16x\(^2\)+16x-12x+12

=3x\(^2\)(x-1)-16x(x-1)-12(x-1)

=(x-1)(3x\(^2\)-16x-12)

=(x-1)(3x\(^2\)-18x+2x-12)

=(x-1)[3x(x-6)+2(x-6)]

=(x-1)(x-6)(3x+2)