Question

bai1:Phân tích đa thức thành nhân tử

a)x^3-x^2-14x+24

b)x^3+4x^2+4x+3

c)x^3-7x-6

d)x^3-19x-30

e)a^3-6a^2+11a-6-

Answer 1

a)\(x^3-x^2-14x+24=x^3-3x^2+2x^2-6x-8x+24\)

\(=x^2\left(x-3\right)+2x\left(x-3\right)-8\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+2x-8\right)\)

=\(\left(x-3\right)\left(x^2+4x-2x-8\right)=\left(x-3\right)\left(x-2\right)\left(x+4\right)\)

b)Tương tự câu a

c)\(x^3-7x+6=x^3-x^2+x^2-x-6x+6\)

=\(x^2\left(x-1\right)+x\left(x-1\right)-6\left(x+1\right)=\left(x+1\right)\left(x^2+x-6\right)\)

=\(\left(x+1\right)\left(x^2+3x-2x-6\right)\)

=\(\left(x+1\right)\left(x-2\right)\left(x+3\right)\)

d;e thương tự câu c