\(A=x^3-x^2-8x+12\)

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

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

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

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

\(A=x^3-x^2-8x+12\)

   \(=x^3-2x^2+x^2-2x-6x+12\)

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

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

   \(=\left(x-2\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)

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

Chúc bạn học tốt.

C1

Ta có \(x^2-8x+12\)

\(=x^2-6x-2x+12\)

\(=x\left(x-6\right)-2\left(x-6\right)\)

\(=\left(x-6\right)\left(x-2\right)\)

C2 

ta có \(x^2-8x+12\)

\(=x^2-8x+16-4\)

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

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

\(=\left(x-6\right)\left(x-2\right)\)

\(x^2-8x+15\)

\(=x^2-3x-5x+15\)

\(=x\left(x-3\right)-5 \left(x-3\right)\)

\(=\left(x-5\right)\left(x-3\right)\)

\(x^2-8x+15\)

\(=\left(x^2-3x\right)-\left(5x-15\right)\)

\(=x\left(x-3\right)-5\left(x-3\right)\)

\(=\left(x-3\right)\left(x-5\right)\)

Tham khảo nhé~

Ta có : x⁴ + 8x² + 7x + 8

= x⁴ - x + 8x² + 8x + 8

= x(x³ - 1) + 8(x² + x + 1)

= x(x - 1)(x² + x + 1) + 8(x² + x + 1)

= (x² - x)(x² + x + 1) + 8(x² + x + 1)

= (x² + x + 1)(x² - x + 8) 

Học tốt nhé !

\(x^4-4x^3+8x^2-16x+16 \)

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

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

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

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

uk cảm ơn góp ý

\(\Leftrightarrow x^3-2x^2+x^2-2x-6x+12\)

\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x-6\right)\)

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

\(\Leftrightarrow\left(x-2\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2\left(x+3\right)\)

T I C K ủng hộ nha 

_________________CHÚC BẠN HỌC TỐT ___________________

Đặt \(x^2-8x+4=t\)

Ta có:\(A=\left(t-2\right)\left(t+2\right)-5=t^2-4-5=t^2-9=\left(t-3\right)\left(t+3\right)\)

Hay \(A=\left(x^2-8x+1\right)\left(x^2-8x+7\right)\)

\(A=\left(x^2-8x+1\right)\left(x-7\right)\left(x-1\right)\)

Cái này nếu phân tích thành nhân tử tiếp thì phức tạp lắm,bạn để như thế này thì gọn hơn

Nếu phân tích ra thì được như thế này:

\(x^2-8x+1=-\left(x+\sqrt{10}+4\right)\left(x+\sqrt{10}-4\right)\)

Bài này giải hệ số bất định.

Ta có:

\(x^4-8x+63\)

\(=x^4+4x^3-4x^3+9x^2-16x^2+7x^2-36x+28x+63\)

\(=\left(x^4-4x^3+7x^2\right)+\left(4x^3-16x^2+28x\right)+\left(9x^2-36x+63\right)\)

\(=x^2\left(x^2-4x+7\right)+4x\left(x^2-4x+7\right)+9\left(x^2-4x+7\right)\)

\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)