`1)x^3-7x+6`

`=x^3-x-6x+6`

`=x(x-1)(x+1)-6(x-1)`

`=(x-1)(x^2+x-6)`

`=(x-1)(x^2-2x+3x-6)`

`=(x-1)[x(x-2)+3(x-2)]`

`=(x-1)(x-2)(x+3)`

`2)x^3-9x^2+6x+16`

`=x^3-2x^2-7x^2+14x-8x+16`

`=x^2(x-2)-7x(x-2)-8(x-2)`

`=(x-2)(x^2-7x-8)`

`=(x-2)(x^2-8x+x-8)`

`=(x-2)[x(x-8)+x-8]`

`=(x-2)(x-8)(x+1)`

`3)x^3-6x^2-x+30`

`=x^3+2x^2-8x^2-16x+15x+30`

`=x^2(x+2)-8x(x+2)+15(x+2)`

`=(x+2)(x^2-8x+15)`

`=(x+2)(x^2-3x-5x+15)`

`=(x+2)[x(x-3)-5(x-3)]`

`=(x+2)(x-3)(x-5)`

`4)2x^3-x^2+5x+3`

`=2x^3+x^2-2x^2-x+6x+3`

`=x^2(2x+1)-x(2x+1)+3(2x+1)`

`=(2x+1)(x^2-x+3)`

`5)27x^3-27x^2+18x-4`

`=27x^3-9x^2-18x^2+6x+12x-4`

`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`

`=(3x-1)(9x^2-6x+4)`

1) Ta có: \(x^3-7x+6\)

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

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

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

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

2) Ta có: \(x^3-9x^2+6x+16\)

\(=x^3-2x^2-7x^2+14x-8x+16\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2-7x-8\right)\)

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

3) Ta có: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

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

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

2x³ + 3x² - 2x

= x ( 2x² + 3x - 2 )

= x ( 2\(x^2\) + 4\(x-x-2\) )

= x [ ( 2\(x^2\) + 4x ) - ( x + 2 )]

= x  [  2x ( x + 2 ) - ( x + 2 )]

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

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

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

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

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

\(8x^4+81\)

\(=8x^4+2\cdot2\sqrt{2}\cdot x^2\cdot9+81-36\sqrt{2}\cdot x^2\)

\(=\left(2\sqrt{2}x^2+9\right)^2-\left(6\sqrt[4]{2}\cdot x\right)^2\)

\(=\left(2\sqrt{2}\cdot x^2-6\sqrt[4]{2}\cdot x+9\right)\left(2\sqrt{2}\cdot x^2+6\sqrt[4]{2}\cdot x+9\right)\)

x⁴ - 2x³ + 2x - 1

= (x⁴ - 1) - (2x³ - 2x)

= (x² - 1)(x² + 1) - 2x(x² - 1)

= (x² - 1)(x² + 1 - 2x)

= (x - 1)(x + 1)(x² - 2x + 1)

= (x - 1)(x + 1)(x - 1)²

= (x - 1)³(x + 1)

Ta có:

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

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

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

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

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

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

5𝑥\(^2\)−17𝑥+6

=5𝑥\(^2\)−2𝑥−15𝑥+6

=𝑥(5𝑥−2)−3(5𝑥−2)

=(𝑥−3)(5𝑥−2)

#Fiona

Chúc bạn học tốt ! 

\(4x^2-5x-6=4x^2-8x+3x-6\)

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

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

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

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

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

Lời giải:

$2x^2-5x-6=2(x^2-\frac{5}{2}x+\frac{5^2}{4^2})-\frac{73}{8}$

$=2(x-\frac{5}{4})^2-\frac{73}{8}$

$=2(x-\frac{5}{4}-\frac{\sqrt{73}}{4})(x-\frac{5}{4}+\frac{\sqrt{73}}{4})$