#)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!

x² - x - 12

= x² + 3x - 4x - 12

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

= (x - 4)(x + 3)

Trả lời:

x² - x - 12 

= x² - 4x + 3x - 12

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

= ( x + 3 ) ( x - 4 )

(x-4)*(x+3)

Phân tích đa thức thành nhân tử

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16\\ =\left(x^2+x+2\right)^2-16\\ =\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\\ =\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

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

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

=\(x^4-x^3+3x^3-3x^2+8x^2+4x-12\)

=\(x^3(x-1)+3x^2(x-1)+4(2x^2+x-3)\)

=\(x^3(x-1)+3x^2(x-1)+4(2x^2-2x+3x-3)\)

=\(x^3(x-1)+3x^2(x-1)+4[2x(x-1)+3(x-1)]\)

=\(x^3(x-1)+3x^2(x-1)+4(x-1)(2x+3)\)

=\((x-1)[x^3+3x^2+4(2x+3)]\)

=\((x-1)(x^3+3x^2+8x+12)\)

\(\left(x^2+x\right)^2+\left(4x^2+4x\right)+4-16\\ =\left(x^2+x+2\right)^2-16\\ =\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)

\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)

=(x-2)(x-6)

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

(x^2+x+1)(x^2+x+2)-12

Đặt x^2+x+1= a ta có

=a^2+a-12

=a^2-3a+4a-12

=(a^2-3a)+(4a-12)

=a(a-3)+4(a-3)

=(a-3)(a+4)

thay x^2+x+1=a ta được

(x^2+x-2)(x^2+x+5)

                                         Bài làm :

Ta có :

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!