\(x^4+4\)

= \(\left(x^2+2\right)^2-4x^2\)

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

Đa thức này không phân tích được thành nhân tử bạn nhé. 

`x^4+x^2 y^2+y^4`

`=x^4+2x^2 y^2 +y^4-x^2 y^2`

`=(x^2+y^2)^2-(xy)^2`

`=(x^2-xy+y^2)(x^2+xy+y^2)`

x⁴ - 2x³-2x² -2x -3

=(x⁴+x³)-(3x³+3x²)+(x²+x)-(3x+3)

=x³(x+1)-3x²(x+1)+x(x+1)-3(x+1)

= (x³-3x²+x-3)(x+1)

= ((x³-3x²)+(x-3))(x+1)

= (x²(x-3)+(x-3))(x+1)

=(x²+1)(x-3)(x+1)

1) =\(x^7-x+x^2+x\)+1

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

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

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

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

=(x^5-x^4+x^2-x)(x^2+x+1)

Trả lời:

1, x⁷ + x² + 1 

= x⁷ + x² + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁷ + x⁶ + x⁵ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁴ + x³ + x² ) - ( x³ + x² + x ) + ( x² + x + 1 )

= x⁵ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x² ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁵ - x⁴ + x² - x + 1 )

b, x⁸ + x⁷ + 1 

= x⁸ + x⁷ + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁸ + x⁷ + x⁶ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁵ + x⁴ + x³ ) - ( x³ + x² + x ) + ( x² + x + 1 ) 

= x⁶ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x³ ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁶ - x⁴ + x³ - x + 1 )

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)