a3+4a2-7a-10

=a3+a2+3a2+3a-10a-10

=a2(a+1)+3a(a+1)-10(a+1)

=(a+1)(a2+3a-10)

=(a+1)(a2-2a+5a-10)

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

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

Phân tích đa thức thành nhân tử

a^3+4a^2-7a-10

a. \(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3\)

Đặt \(a+b-c=x\) , \(b+c=y,c+a-b=z\) thì:

\(x+y+z=a+b-c+b+c-a+c+a-b=a+b+c\)

Áp dụng hằng đẳng thức ta có:

\(\left(x+y+z\right)^3-x^3-y^3-z^3=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Ta có: \(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3\left(c+a-b\right)^3\)

\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\)

\(=3.2b.2c.2a=24abc\)

b. \(abc-\left(ab+bc+ca\right)\left(a+b+c\right)-1\)

\(=abc-bc-ab+b-ac+c+a-1\)

\(=bc\left(a-1\right)-b\left(a-1\right)-c\left(a-1\right)+\left(a-1\right)\)

\(=\left(a-1\right)\left(bc-b-c+1\right)=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)