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a) x3 - 7x + 6
= x3 - 2x2 + 2x2 - 4x - 3x + 6
= x2 ( x - 2 ) + 2x ( x - 2 ) - 3 ( x - 2 )
= ( x - 2 ) ( x2 + 2x - 3 )
= ( x - 2 ) ( x2 - x + 3x - 3 )
= ( x - 2 ) [ x ( x - 1 ) + 3 ( x - 1 ) ]
= ( x - 2 ) ( x - 1 ) ( x + 3 )
b ) x3 - 9x2 + 6x + 16
= x3 - 8x2 - x2 + 8x - 2x + 16
= x2 ( x - 8 ) - x ( x - 8 ) - 2 ( x - 8 )
= ( x - 8 ) ( x2 - x - 2 )
= ( x - 8 ) ( x2 + x - 2x - 2 )
= ( x - 8 ) [ x ( x + 1 ) - 2 ( x + 1 ) ]
= ( x - 8 ) ( x + 1 ) ( x - 2 )
c ) x3 - 6x2 - x + 30
= x3 - 5x2 - x2 + 5x - 6x + 30
= x2 ( x - 5 ) - x ( x - 5 ) - 6 ( x - 5 )
= ( x - 5 ) ( x2 - x - 6 )
= ( x - 5 ) ( x2 - 3x + 2x - 6 )
= ( x - 5 ) [ x ( x - 3 ) + 2 ( x - 3 ) ]
= ( x - 5 ) ( x - 3 ) ( x + 2 )
d ) 2x3 - x2 + 5x + 3
= 2x3 + x2 - 2x2 - x + 6x + 3
= x2 ( 2x + 1 ) - x ( 2x + 1 ) + 3 ( 2x + 1 )
= ( 2x + 1 ) ( x2 - x + 3 )
a) \(x^2-x-2=x^2+x-2x-2=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
a) \(x^2-x-2=x^2-2x+x-2=x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x+1\right)\)
b) \(x^3-19x-30==x^3+2x^2-2x^2-4x-15x-30=x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x-15\right)=\left(x+2\right)\left(x-3\right)\left(x+5\right)\)
c) \(x^3-6x^2+11x-6=x^3-x^2-5x^2+5x+6x-6=x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
ta có \(x^3+x^2-x+2=x^3+2x^2-x^2-2x+x+2=x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)=\left(x+2\right)\left(x^2-x+1\right)\)
b)ta có \(x^3-6x^2-x+30=x^3+2x^2-8x^2-16x+15x+30\)
\(=\left(x+2\right)\left(x^2-8x+15\right)=\left(x+2\right)\left[\left(x^2-3x\right)-\left(5x-15\right)\right]=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
a]
x^3 + 6x^2 + 11x + 6
= x^3 + x^2 + 5x^2 + 5x + 6x + 6
= x^2(x + 1) + 5x(x + 1) + 6(x + 1)
= (x + 1)(x^2 + 5x + 6)
= (x + 1)(x^2 + 2x + 3x + 6)
= (x + 1)[x(x + 2) + 3(x + 2)
= (x + 1)(x + 2)(x + 3)
b]
x^3 + 8x^2 + 19x + 12
= x^3 + x^2 + 7x^2 + 7x + 12x + 12
= x^2(x + 1) + 7x(x + 1) + 12(x + 1)
= (x + 1)(x^2 + 7x + 12)
= (x + 1)(x^2 + 3x + 4x + 12)
= (x + 1)[x(x + 3) + 4(x + 3)]
= (x + 1)(x + 3)(x + 4)
\(^{x^3-6x^2-x+30=x^3-5x^2-3x^2+15x-2x^2-10x-6x+30}\)
=x^2(x-5)-3x(x-5)-2x(x-5)-6(x-5)
=(x-5)(x^2-3x-2x-6)
=(x-5)[x(x-3)-2(x-3)]
=(x-5)(x-3)(x-2)
\(x^3-6x^2-x+30\)
= \(x^3-5x^2-3x^2+15x+2x^2-10x-6x+30\)
= \(x^2\left(x-5\right)-3x\left(x-5\right)+2x\left(x-5\right)-6\left(x-5\right)\)
= \(\left(x-5\right)\left(x^2-3x+2x-6\right)\)
= \(\left(x-5\right)\left(x\left(x-3\right)+2\left(x-3\right)\right)\)
= \(\left(x-5\right)\left(x+2\right)\left(x-3\right)\)
a)\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x^2-5x-3x+15\right)\)
\(=\left(x+2\right)\left[x\left(x-5\right)-3\left(x-5\right)\right]\)
\(=\left(x+2\right)\left(x-5\right)\left(x-3\right)\)
nha
a. \(-x^3-6x^2+6x+1=-x^3+x^2-7x^2+7x-x+1=\left(1-x\right)\left(x^2+7x+1\right)\)
b. \(x^4-4x^2+4x-1=x^4-1-4x\left(x-1\right)=\left(x-1\right)\left[\left(x+1\right)\left(x^2+1\right)-4x\right]\)
\(=\left(x-1\right)\left(x^3+x^2-3x+1\right)\)
c. \(6x^3-x^2-486x+81=6x^3-54x^2+53x^2-477x-9x+81=\left(x-9\right)\left(6x^2+53x-9\right)\)
\(=\left(x-9\right)\left(x+9\right)\left(6x-1\right)\)
d. \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)=x^2\left(x^2+8x+16\right)-x^2-8x-16-x^2+1\)
\(=x^4+8x^3+14x^2-8x-15=x^4+5x^3+3x^3+15x^2-x^2-5x-3x-15\)
\(=\left(x+5\right)\left(x^3+3x^3-x-3\right)=\left(x+5\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
Để phân tích nhân tử các dạng này, em cần nhẩm được nghiệm để biết đc nhân tử chung là gì, sau đó tách để xuất hiện nhân tử chung đó. CHÚC EM HỌC TỐT :))