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

#)Giải :

\(x^3-2x-4\)

\(=x^3+2x^2-2x^2+2x-4x-4\)

\(=x^3+2x^2+2x-2x^2-4x-4\)

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

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

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

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

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

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

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

\(=x^3-2x^2+2x^2-4x+2x-4\)

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

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

cảm ơn nha!

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

(x-4)*(x+3)

Phân tích đa thức thành nhân tử

\(x^2-x-12=x^2-x-9-3\)

\(=\left(x^2-9\right)-\left(x+3\right)\)

\(=\left(x-3\right).\left(x+3\right)-\left(x+3\right)\)

\(=\left(x+3\right).\left(x-4\right)\)

\(=3x^2+6x-14x+2x^2\\ =5x^2-8x=x\left(5x-8\right)\)

Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

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

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

\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)

\(x^2-x-12=x^2-3x+4x-12\)

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

\(=x\left(x-3\right)+4\left(x-3\right)\)

\(=\left(x-3\right)\left(x+4\right)\)

\(x^2-x-12=x^2-4x+3x-12\)

\(=x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

222+x)+(3x+3)=x(x+1)+3(x+1)=(x+3)(x+1)

Trả lời:

x² + 4x + 3

= ( x² + 2.x.2 + 2² ) - 1

= ( x + 2 )² - 1

= ( x + 2 - 1 ) ( x + 2 + 1 )

= ( x + 1 ) ( x + 3 )