=> x (x³ + 5x² - 7x - 41)

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

đề sai ko? ~0~

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) \(x^6-14x^4+49x^2-36\)

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

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

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

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

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

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

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

= x^4  + x^3  + 4x^3 + 4x^2 - 11x^2  - 11x -30x - 3 

= x^3 ( x + 1) + 4x^2 ( x + 1)- 11x( x + 1) - 30 ( x+ 1)

= ( x + 1)(x^3 + 4x^2 - 11x - 30)

=  ( x + 1) [ x^3 - 3x^2 + 7x^2 - 21x + 10x - 30]

= ( x  + 1) [ x^2 ( x - 3) + 7x(x - 3) + 10 ( x- 3) ] 

= ( x + 1 ) (x - 3 )(x^2 + 7x + 10)

= ( x + 1)(x - 3) [ x^2 + 2x + 5x + 10)

= ( x+ 1)(x-3) [ x(x+2) + 5(x+2) ] 

=(x + 1)(x-3)(x+2)(x+5)

khong co bai nao la kho chi la ko biet lam

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

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

= (x+1).(x³ + 4x² -11x - 30) = (x+1).(x³ - 3x² + 7x² - 21x + 10x - 30)

= (x+1).[x².(x - 3) + 7x.(x - 3) + 10.(x - 3)] = (x +1).[(x - 3).(x² + 7x + 10)] = (x+1).(x - 3). (x² + 2x + 5x + 10) 

= (x + 1).(x - 3).(x + 2).(x + 5) 

6x³ + 5x² - 7x - 4

= (6x³ + 5x - 7x) - 4

= x (6x² - 5 - 7) - 2²

= x (6x² - 12) - 2²

= x [6 (x² - 2)] - 2²

= x [6 (x² - \(\sqrt{2}^2\))] - 2²

= x [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))] - 2²

= (x - 2²) [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))

b) 2x³ - x² + x - 2

= (2x³ - x² - x) - 2

= x (2x² - x - 1) - 2

= (x - 2) (2x² - x - 1)

(mik ko biet dug ko, neu sai mog bn thog cam)