Phân tích đa thức thành nhân tử:
\(x^3-6x^2-x+30\)
\(x^3-x^2-x-2\)
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KM
1
25 tháng 8 2017
\(\Leftrightarrow x^3-3x^2-3x^2+9x-10x+30\)
\(\Leftrightarrow x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x-10\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)\left(x+2\right)\)
HN
1
TM
18 tháng 8 2017
\(=\left(x^3-2x^2\right)+\left(8x^2-16x\right)+\left(15x-30\right)\)
\(=x^2\left(x-2\right)+8x\left(x-2\right)+15\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+8x+15\right)\)
\(=\left(x-2\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x-2\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x-2\right)\left(x+3\right)\left(x+5\right)\)
SM
1 tháng 10 2018
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 )
TM
2
23 tháng 6 2016
\(^{x^3-6x^2-x+30=x^3-5x^2-3x^2+15x-2x^2-10x-6x+30}\)
=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)
27 tháng 10 2019
\(x^3-6x^2-x+30\)
= \(x^3-5x^2-3x^2+15x+2x^2-10x-6x+30\)
= \(x^2\left(x-5\right)-3x\left(x-5\right)+2x\left(x-5\right)-6\left(x-5\right)\)
= \(\left(x-5\right)\left(x^2-3x+2x-6\right)\)
= \(\left(x-5\right)\left(x\left(x-3\right)+2\left(x-3\right)\right)\)
= \(\left(x-5\right)\left(x+2\right)\left(x-3\right)\)
TM
1
1 tháng 7 2016
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
30 tháng 9 2021
e) \(=x^2\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^2-2x+3\right)\)
g) \(=x^2\left(3x-1\right)-x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(x^2-x+4\right)\)
h) \(=3x^2\left(2x+1\right)-x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)
i) \(=2x^2\left(2x+1\right)+2x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(2x^2+2x+1\right)\)
DM
4
TN
19 tháng 6 2016
x3-6x2-x+30
=x3-5x2-x2+5x-6x+30
=(x-5)(x2-x-6)
=(x-5)(x-3)(x+2)
HT
19 tháng 6 2016
\(x^3-6x^2-x+30=x^3-3x^2-3x^2+9x-10x+30.\)
\(=x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-3x-10\right)\)
\(=\left(x-3\right)\left(x^2+2x-5x-10\right)\)
\(=\left(x-3\right)\left[x\left(x+2\right)-5\left(x+2\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)
Vậy \(x^3-6x^2-x+30=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)
VB
1
TN
30 tháng 8 2016
\(2x^4+x^3-6x^2+x+2\)
= \(2x^4+4x^3-3x^3-6x^2+x+2\)
= \(2x^3\left(x+2\right)-3x^2\left(x+2\right)+\left(x+2\right)\)
= \(\left(x+2\right)\left(2x^3-3x^2+1\right)\)
=\(\left(x+2\right)\left(2x^3-2x^2-x^2+1\right)\)
=\(\left(x+2\right)\left(2x^2\left(x-1\right)-\left(x+1\right)\left(x-1\right)\right)\)
=\(\left(x+2\right)\left(x-1\right)\left(2x^2-x-1\right)\)
= \(\left(x+2\right)\left(x-1\right)\left(2x^2-2x+x-1\right)\)
=\(\left(x+2\right)\left(x-1\right)\left(2x\left(x-1\right)+\left(x-1\right)\right)\)
=\(\left(x+2\right)\left(2x+1\right)\left(x-1\right)^2\)
VT
14 tháng 9 2017
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 \(x^3-6x^2-x+30=x^3+2x^2-8x^2-16x+15x+30\)
\(=\left(x+2\right)\left(x^2-8x+15\right)=\left(x+2\right)\left[\left(x^2-3x\right)-\left(5x-15\right)\right]=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
14 tháng 9 2017
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)
k trc làm sau sẵn kb lun
k trc làm sau đấy