Question

Phân tích thành nhân tử : 

x³+4x²-7x-10

Answer 1

\(x^3+4x^2-7x-10\)

\(=x^3-2x^2+6x^2-12x+5x-10\)

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

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

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

\(=\left(x-2\right)\left[x\left(x+1\right)+5\left(x+1\right)\right]\)

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

Answer 2

x³+4x²-7x-10

= x³-2x²+6x²-12x+5x-10

= x²(x-2)+6x(x-2)+5(x-2)

= (x²+6x+5)(x-2)

= (x²+x+5x+5)(x-2)

= [x(x+1)+5(x+1)](x-2)

= (x+5)(x+1)(x-2)