a.\(3x^3-x^2-21x+7=\)\(x^2\left(3x-1\right)-7\left(3x-1\right)=\left(3x-1\right)\left(x^2-7\right)\)

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

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

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

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

Bạn xem lại nhé. Đa thức đã cho không phân tích được thành nhân tử nữa. Chỉ thu gọn được thành $8x^3+50x^2-55x+10$

\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)

a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)

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

c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)

d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)

e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)

\(\text{e) 2x^4 + 5x^3+13x^2+25x+15 }\)
  \(\text{=2x^3(x+1)+3x^2(x+1)+10x(x+1)+15(x+1) }\)
  \(\text{=(x+1)[x^2(2x+3)+5(2x+3)]}\)
  \(\text{=(x+1)(2x+3)(x^2+5)}\)

\(2x^4+5x^3+13x^2+25x+15\)

\(=2x^4+2x^3+3x^3+3x^2+10x^2+10x+15x+15\)

\(=2x^3\left(x+1\right)+3x^2\left(x+1\right)+10x\left(x+1\right)+15\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^3+3x^2+10x+15\right)\)

Làm tiếp

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

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

1. 5x² - 4x

= x(5x - 4)

2. 8x²(x - 3y) - 12x(x - 3y)

= (8x² - 12x)(x - 3y)

= 4x(2x - 3)(x - 3y)

3. 3(x - y) - 5x(y - x)

= 3(x - y) + 5x(x - y)

= (3 + 5x)(x - y)