$D=\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+15$

$D=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+15$

$D=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+15$

Đặt x2 - 10x + 20 = a, ta được

$D=\left(a-4\right)\left(a+4\right)+15$

$D=a^2-16+15$

$D=a^2-1$

$D=\left(a-1\right)\left(a+1\right)$

$D=\left(x^2-10x+20-1\right)\left(x^2-10x+20+1\right)$

$D=\left(x^2-10x+19\right)\left(x^2-10x+21\right)$

$D=\left(x^2-10x+19\right)\left(x^2-3x-7x+21\right)$

$D=\left(x^2-10x+19\right)\left[x\left(x-3\right)-7\left(x-3\right)\right]$

$D=\left(x^2-10x+19\right)\left(x-3\right)\left(x-7\right)$

phân tích đa thức thành nhân tử:$x^3+5x^2+3x-9$

$x^3+5x^2+3x-9$

$=x^3+3x^2+2x^2+6x-3x-9$

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

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

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

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

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

Phan tich da thuc sau thanh nhan tu :

$6x^2-11x+3$

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

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

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