Phân tích đa thức thành nhân tử
1, ,x8+x+1
2,x4+2008x2+2007x+2008
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NH
2
H
6 tháng 4 2021
\(x^4+2008x^2+2007x+2008\\ =x^4-x+2008\left(x^2+x+1\right)=x\left(x^3-1\right)+2008\left(x^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
6 tháng 4 2021
Ta có: \(x^4+2008x^2+2007x+2008\)
\(=x^4-x+2008\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
M
0
H
2
7 tháng 2 2018
mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
D
7 tháng 2 2018
x^4 + 2008x^2 + 2007x + 2008
\(=x^4+x^2+2007x^2+2007x+2007+1\)
\(=x^4+x^2+1+2007\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
VM
1
26 tháng 8 2021
a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b: \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
c: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
VA
1
14 tháng 2 2016
x^4+2008x^2+2007x+2008
=x^4+2008x^2+2008x-x+2008
=(x^4-x)+(2008x^2+2008x+2008)
=x(x^3-1)+2008(x^2+x+1)
=x(x-1)(x^2+x+1)+2008(x^2+x+1)
=(x^2+x+1)(x^2-x+2008)
NT
1
18 tháng 6 2018
x4+2008x2+2007x+2008
<=> x4-x+2008x2+2008x+2008
<=> x(x3-1)+2008(x2+x+1)
<=> x(x-1)(x2+x+1)+2008(x2+x+1)
<=> (x2+x+1)(x2-x+2008)
VT
2
14 tháng 3 2015
\(\left(x^4+x^2+1\right)+\left(2007x^2+2007x+2007\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
NL
2
21 tháng 8 2021
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
TP
2
21 tháng 8 2021
1) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
3) \(x^2+7x+12=\left(x^2+3x\right)+\left(4x+12\right)=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
21 tháng 8 2021
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
a) \(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
b) \(x^4+2008x^2+2007x+2008\)
\(=x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
\(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)