\(1,\)

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

Đặt: \(\left(x^2-x+2\right)^2=n\)

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

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

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

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

Thay \(\left(x^2-x+2\right)^2\)ta có:

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

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

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

\(2,\)

\(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4\)

\(=3\left(-x^2+2x+3\right)^4+x^2\left(-x^2+2x+3\right)^2-27x^2\left(-x^2+2x+3\right)^2-9x^4\)

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

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

\(x^2+2xy+y^2+2x+2y-15\)

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

\(=\left(x+y+1\right)^2-4^2=\left(x+y-3\right)\left(x+y+5\right)\)

\(x^2+2xy+y^2+2x+2y-15\)

\(=x^2+2xy+y^2+2x+2y+1-16\)

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

Đặt \(x+y=t\)

\(\Rightarrow t^2+2t+1-16\)

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

\(=\left(t+1-4\right)\left(t+1+4\right)\)

\(=\left(t-3\right)\left(t+5\right)\)

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

phân tích đa thức thành nhân tử 4x^3+6x^2+4x+1

= 4x^3+6x^2+4x+1

= (2x+1)(2x^2+2x+1)

nha bạn chúc bạn học tốt nha 

\(4x^3+6x^2+4x+1\)

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

\(=\left(4x^3+4x^2+2x\right)+\left(2x^2+2x+1\right)\)

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

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

phân tích đa thức thành nhân tử 6x^3+x^2+x+1

= 6x^3+x^2+x+1

= (2x+1)(3x^2-x+1)

chúc bạn học tốt nha 

6x³ + x² + x+1

=6x³ +3x² - 2x² - x + 2x +1

= 3x²(2x + 1) - x(2x + 1) + (2x+1)

=(2x+1)(3x² -x +1) 

\(x^3+3x^2-10x-24\)

\(=x^3-3x^2+6x^2-18x+8x-24\)

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

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

\(=\left(x-3\right)\left(x^2+6x+9-1\right)\)

\(=\left(x-3\right)[\left(x-3\right)^2-1]\)

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

\(2x^3-11x^2+10x+8\)

\(=2x^3-4x^2-7x^2+14x-4x+8\)

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

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

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

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

Ta có: \(^{3x^3-4x^2+13x-4}\) = \(3x^3-x^2-3x^2+x+12x-4\)

                                       = \(3x^2\left(x-\frac{1}{3}\right)-3x\left(x-\frac{1}{3}\right)+12\left(x-\frac{1}{3}\right)\)

                                       = \(\left(3x^2-3x+12\right)\left(x-\frac{1}{3}\right)\)

                                       = \(3\left(x^2-x+4\right)\left(x-\frac{1}{3}\right)\)

 3x^3-4x^2+13x-4

= (3x-1)(x^2-x+4)

nha bạn 

\(2x^3-35x+75=2x^3+10x^2-10x^2-50x+15x+75\)

\(=\left(x+5\right)\left[2x^2-10x+15\right]\)

Trả lời:

Sửa đề: \(2x^2-35x+75\)

\(=2x^2-30x-5x+75\)

\(=\left(2x^2-30x\right)-\left(5x-75\right)\)

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

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

\(1,2x^3+3x^2-8x+3\)

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

\(=2x^2\left(x-1\right)+5x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(2x^2+5x-3\right)\left(x-1\right)\)

\(=\left(2x-1\right)\left(x+3\right)\left(x-1\right)\)

\(2,x^3-5x^2+2x+8\)

\(=x^3+x^2-6x^2-6x+8x+8\)

\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)

\(=\left(x^2-6x+8\right)\left(x+1\right)\)

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

\(3,-6x^3+x^2+5x-2\)

\(=-6x^3-6x^2+7x^2+7x-2x-2\)

\(=-6x^2\left(x+1\right)+7x\left(x+1\right)-2\left(x+1\right)\)

\(=\left(-6x^2+7x-2\right)\left(x+1\right)\)

\(=\left(-6x^2-3x-4x-2\right)\left(x+1\right)\)

\(=\left[-3x\left(2x+1\right)-2\left(2x+1\right)\right]\left(x+1\right)\)

\(=\left(-3x-2\right)\left(2x+1\right)\left(x+1\right)\)

\(4,3x^3+19x^2+4x-12\)

\(=3x^3+18x^2+x^2+6x-2x-12\)

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

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

\(=\left(3x-2\right)\left(x+1\right)\left(x+6\right)\)