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

\(=\left(a^3+5a^2\right)-\left(a^2+5a\right)-\left(2a+10\right)\)

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

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

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

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

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

tự làm di

dễ đều có nhân tử chung là (x+1)

b. \(\left(a^2+a\right)+a\left(a^2+a\right)-12\)

<=>\(\left(x^3+3x^2-4\right)+\left(3x^2+9x-12\right)\)

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

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

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

đóngmở ngoặc nhé mk ngại ghi lại

<=>(x+3)(x(x+4)-(x+4))

<=>(x+3)(x-1)(x+4)

kết pn fb mk nhé longtrangv@gmail.com

c) \(x^3-x^2-4x^2+8x-4\)

= \(x^3-x^2-4x^2+4x+4x-4\)

= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)

= \(\left(x-1\right)\left(x^2-4x+4\right)\)

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

\(=7\left(a-b\right)+\left(a-b\right)\left(a+b\right)=\left(a-b\right)\left(7+a+b\right)\)

\(7a-7b+a^2-b^2\)

\(=7\left(a-b\right)+\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(a+b+7\right)\)

\(a^3+4a^2+4a+3\)

\(=a^3+a^2+3a^2+3a+a+3\)

\(=\left(a^3+a^2+a\right)+\left(3a^2+3a+3\right)\)

\(=a\left(a^2+a+1\right)+3\left(a^2+a+1\right)\)

\(=\left(a+3\right)\left(a^2+a+1\right)\)

sai đề huj