Question

phân tích đa thức thành nhân tử

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

Answer 1

x³+4x²-7x-10=x³+x²+3x²+3x-10x-10

=x²(x+1)+3x(x+1)-10(x+1)=(x+1)(x²+3x-10)=(x+1)[x²+5x-2x-10]

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

Answer 2

x3+4x2−7x−10 = \(x^3+x^2+3x^2+3x-10x-10\)

=x²(x+1)+3x(x+1)-10(x+1)=(x+1)(x²+3x-10)=(x+1)[x²+5x-2x-10]

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