Như thế này : 

x^3 - 8x^2 + x + 42 = x^3 - 7x^2 - x^2 + 7x - 6x + 42

                               = ( x^3 - x^2 ) - ( 7x^2 - 7x ) - ( 6x - 42 )

                               = x^2.( x - 1 ) - 7x.( x - 1 ) - 6.( x - 7 )

                               = ( x^2 - 7x ).( x - 1 ) - 6.( x - 7 )

                               = x.( x- 7 ).( x - 1 ) - 6.( x - 7 )  =  [ x.( x - 1 ) - 6 ].( x - 7 ) 

x^4 + 5x^3 - 7x^2 - 41x - 30 = x^4 + 5x^3 - 7x^2 - 35x - 6x - 30

                                            = x.( x^3 + 6 ) + 5.( x^3 + 6 ) - 7x.( x + 5 )

                                            = ( x + 5 ) ( x^3 + 6 ) - 7x.( x + 5 ) 

                                            = ( x + 5 ).( x^3 - 7x + 5 )

CHÚC BẠN HỌC TỐT

=> x (x³ + 5x² - 7x - 41)

a) \(x^4+3x^3-7x^2-27x-18\)

\(=\left(x^4+3x^3+2x^2\right)-\left(9x^2-27x-18\right)\)

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

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

b) \(x^4+5x^3-7x^2-41x-30\)

\(=\left(x^4+2x^3-15x^2\right)+\left(3x^3+6x^2-45x\right)+\left(2x^2+4x-30\right)\)

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

\(=\left(x^2+2x-15\right)\left(x^2+3x+2\right)=\left(x^2+5x-3x-15\right)\left(x^2+x+2x+2\right)\)

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

c) \(x^6-14x^4+49x^2-36\)

\(=\left(x^6-9x^4\right)+\left(-5x^4+45x^2\right)+\left(4x^2-36\right)\)

\(=x^4\left(x^2-9\right)-5x^2\left(x^2-9\right)+4\left(x^2-9\right)\)

\(=\left(x^2-9\right)\left(x^4-5x^2+4\right)=\left(x^2-9\right)\left(x^4-4x^2-x^2+4\right)\)

\(=\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)

c: \(x^3-8x^2+x+42\)

\(=x^3+2x^2-10x^2-20x+21x+42\)

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

\(=\left(x+2\right)\left(x-3\right)\left(x-7\right)\)

a: \(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

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

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

Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]       
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
 

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

2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)

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

5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)

6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)

1, x³+ 6x²+11x+6

= x³ + 2x² + 4x² + 8x + 3x + 6 

= x²(x + 2) + 4x(x + 2) + 3(x + 2)

= (x + 2)(x² + 4x + 3)

2, x⁴+3x³-7x²-27x-18

= x⁴ + 3x³ - 9x² + 2x² - 27x -18

= (x⁴ - 9x²) + (3x³ - 27x) + (2x² - 18)

= x²(x² - 9) + 3x(x² - 9) + 2(x² - 9)

= (x² - 9)(x² + 3x + 2)

= (x + 3)(x - 3)(x² + 3x + 2)

3, x³-8x²+x+42

= x³ - 7x² - x² + 7x - 6x + 42

= (x³ - 7x²) - (x² - 7x) - (6x - 42)

= x²(x - 7) - x(x - 7) - 6(x - 7)

= (x - 7)(x² - x - 6) 

4, x⁴+5x³-7x²-41x-30 

= x⁴ + x³ + 4x³ - 4x² - 11x² - 11x - 30x - 30

= (x⁴ + x³) + (4x³ - 4x²) - (11x² + 11x) - (30x + 30)

= x³(x + 1) + 4x²(x + 1) - 11x(x + 1) - 30(x + 1)

= (x³ + 4x² - 11x - 30)(x + 1)

5, x⁵+x-1

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

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

= (x² - x + 1)(x³ + x² - 1)

6, x⁵-x⁴-1

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

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

= (x² - x + 1)(x³ - x - 1)

1, x 3+ 6x 2+11x+6

= x 3 + 2x 2 + 4x 2 + 8x + 3x + 6

= x 2 ﴾x + 2﴿ + 4x﴾x + 2﴿ + 3﴾x + 2﴿

= ﴾x + 2﴿﴾x 2 + 4x + 3﴿

2, x 4+3x 3‐7x 2‐27x‐18

= x 4 + 3x 3 ‐ 9x 2 + 2x 2 ‐ 27x ‐18

= ﴾x 4 ‐ 9x 2 ﴿ + ﴾3x 3 ‐ 27x﴿ + ﴾2x 2 ‐ 18﴿

= x 2 ﴾x 2 ‐ 9﴿ + 3x﴾x 2 ‐ 9﴿ + 2﴾x 2 ‐ 9﴿

= ﴾x 2 ‐ 9﴿﴾x 2 + 3x + 2﴿

=﴾x + 3﴿﴾x ‐ 3﴿﴾x 2 + 3x + 2﴿

3, x 3‐8x 2+x+42

= x 3 ‐ 7x 2 ‐ x 2 + 7x ‐ 6x + 42

= ﴾x 3 ‐ 7x 2 ﴿ ‐ ﴾x 2 ‐ 7x﴿ ‐ ﴾6x ‐ 42﴿

= x 2 ﴾x ‐ 7﴿ ‐ x﴾x ‐ 7﴿ ‐ 6﴾x ‐ 7﴿

= ﴾x ‐ 7﴿﴾x 2 ‐ x ‐ 6﴿

4, x 4+5x 3‐7x 2‐41x‐30

= x 4 + x 3 + 4x 3 ‐ 4x 2 ‐ 11x 2 ‐ 11x ‐ 30x ‐ 30

= ﴾x 4 + x 3 ﴿ + ﴾4x 3 ‐ 4x 2 ﴿ ‐ ﴾11x 2 + 11x﴿ ‐ ﴾30x + 30﴿

= x 3 ﴾x + 1﴿ + 4x 2 ﴾x + 1﴿ ‐ 11x﴾x + 1﴿ ‐ 30﴾x + 1﴿

= ﴾x 3 + 4x 2 ‐ 11x ‐ 30﴿﴾x + 1﴿

5, x 5+x‐1

= x 5 ‐ x 4 + x 3 + x 4 ‐ x 3 + x 2 ‐ x 2+ x ‐1

= x 3 ﴾x 2 ‐ x + 1﴿+ x 2 ﴾x 2 ‐ x + 1﴿‐ ﴾x 2 ‐ x + 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 + x 2 ‐ 1﴿ 6, x 5‐x 4‐1

= x 5 ‐ x 3 ‐ x 2 ‐ x 4 + x 2 + x + x 3 ‐ x ‐ 1

= x 2 ﴾x 3 ‐ x ‐ 1﴿ ‐ x﴾x 3 ‐ x ‐ 1﴿ + ﴾x 3 ‐ x ‐ 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 ‐ x ‐ 1﴿ 

a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)

\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)

\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)

\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)

\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)

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

b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)

\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)

\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)

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

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

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