phân tích thành nhân tử
\(6x^4+5x^3-38x^2+5x+6\)
K
Khách
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T
0
16 tháng 8 2020
a) \(=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x^3-2x^2+2x-1\right)\left(x-1\right)\)
\(=\left(x^3-x^2-x^2+x+x-1\right)\left(x-1\right)\)
\(=\left(x^2-x+1\right)\left(x-1\right)^2\)
c)
\(=6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6\)
\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)
\(=\left(6x^3+18x^2-x^2-3x-x-3\right)\left(x-2\right)\)
\(=\left(6x^2-x-1\right)\left(x+3\right)\left(x-2\right)\)
\(=\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\left(x-2\right)\)
16 tháng 8 2020
b)
\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)
\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)
\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)
CÓ: \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)
=> \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)
=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.
BN
1
KN
13 tháng 9 2019
a)\(x^4+6x^3+11x^2+6x+1\)
\(=x^4+9x^2+1+6x^3+6x+2x^2\)
\(=\left(x^2+3x+1\right)^2\)
EC
3
SM
19 tháng 9 2018
A = 6x4 - 5x3 + 4x2 + 2x - 1
= 6x4 + 3x3 - 8x3 - 4x2 + 8x2 + 4x - 2x - 1
= 3x3. ( 2x + 1 ) - 4x2 ( 2x + 1 ) + 4x ( 2x + 1 ) - ( 2x + 1 )
= ( 2x + 1 ) ( 3x3 - 4x2 + 4x - 1 )
= ( 2x + 1 ) ( 3x3 - x2 - 3x2 + x + 3x - 1 )
= ( 2x + 1 ) [ x2 ( 3x - 1 ) - x ( 3x - 1 ) + ( 3x - 1 ) ]
= ( 2x + 1 ) ( 3x - 1 ) ( x2 - x + 1 )
B = 4x4 + 4x3 + 5x2 + 8x - 6
= 4x4 - 2x3 + 6x3 - 3x2 + 8x2 - 4x + 12x - 6
= 2x3 ( 2x - 1 ) + 3x2 ( 2x - 1 ) + 4x ( 2x - 1 ) + 6 ( 2x - 1 )
= ( 2x - 1 ) ( 2x3 + 3x2 + 4x + 6 )
= ( 2x - 1 ) [ x2 ( 2x + 3 ) + 2 ( 2x + 3 ) ]
= ( 2x - 1 ) ( 2x + 3 ) ( x2 + 2 )
C = x4 + x3 - 5x2 + x - 6
= x4 - 2x3 + 3x3 - 6x2 + x2 - 2x + 3x - 6
= x3 ( x - 2 ) + 3x2 ( x - 2 ) + x ( x - 2 ) + 3 ( x - 2 )
= ( x - 2 ) ( x3 + 3x2 + x + 3 )
= ( x - 2 ) [ x2 ( x + 3 ) + ( x + 3 ) ]
= ( x - 2 ) ( x + 3 ) ( x2 + 1 )
TN
2
KN
4 tháng 1 2021
5x3 + 38x2 + 19x - 14
= ( 5x3 + 35x2 ) + ( 3x2 + 21x ) - ( 2x + 14 )
= 5x2 ( x + 7 ) + 3x ( x + 7 ) - 2 ( x + 7 )
= ( x + 7 ) ( 5x2 + 3x - 2 )
= ( x + 7 ) [ ( 5x2 - 2x ) + ( 5x - 2 ) ]
= ( x + 7 ) [ x ( 5x - 2 ) + ( 5x - 2 ) ]
= ( x + 7 ) ( x + 1 ) ( 5x - 2 )
NC
5 tháng 1 2021
\(5x^3+38x^2+19x-4\)
\(=\left(5x^3+35x^2\right)+\left(3x^2+21x\right)-\left(2x+14\right)\)
\(=5x^2\left(x+7\right)+3x\left(x+7\right)-2\left(x+7\right)\)
\(=\left(5x^2+3x-2\right)\left(x+7\right)\)
\(=\left(5x^2-2x+5x-2\right)\left(x+7\right)\)
\(=\left[x\left(5x-2\right)+\left(5x-2\right)\right]\left(x+7\right)\)
\(=\left(x+1\right)\left(5x-2\right)\left(x+7\right)\)
\(6x^4+5x^3-38x^2+5x+6\)
= \(6x^4-15x^3+6x^2+20x^3-50x^2+20x+6x^2-15x+6\)
= \(3x^2\left(2x^2-5x+2\right)+10x\left(2x^2-5x+2\right)+3\left(2x^2-5x+2\right)\)
= \(\left(2x^2-5x+2\right)\left(3x^2+10x+3\right)\)
= \(\left(2x^2-x-4x+2\right)\left[3x^2+x+9x+3\right]\)
= \(\left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\)
= \(\left(x-2\right)\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\)
Chúc bạn học tốt !!!