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\(\Leftrightarrow x^3-3x^2-3x^2+9x-10x+30\)
\(\Leftrightarrow x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x-10\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x^3-2x^2\right)+\left(8x^2-16x\right)+\left(15x-30\right)\)
\(=x^2\left(x-2\right)+8x\left(x-2\right)+15\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+8x+15\right)\)
\(=\left(x-2\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x-2\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x-2\right)\left(x+3\right)\left(x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt y=x2+1
=>y2=(x2+1)2
y2=x4+2x2+1
đặt P(x)=x^4+6x^3+11x^2+6x+1
=x4+2x2+1+6x3+6x+9x2
=x4+2x+1+6x(x2+1)+9x2
thay y2=x4+2x2+1 và y=x2+1 ta được
Q(y)=y2+6xy+9x2
=(y+3x)2
thay y=x2+1 ta được:
(x2+3x+1)2
vậy x^4+6x^3+11x^2+6x+1=(x2+3x+1)2
![](https://rs.olm.vn/images/avt/0.png?1311)
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= (1 - x3 ) + ( 6x - 6x2 )
= (1 - x ).(1 + x + x2) + 6x.(1 - x)
= (1 - x).(1+x+x2 + 6x)
= (1 - x).(1 + 7x +x2 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x\left(2x^2-x-6\right)\)
\(=x\left(2x^2-4x+3x-6\right)\)
\(=x\left[2x\left(x-2\right)+3\left(x-2\right)\right]\)
\(=x\left(x-2\right)\left(2x+3\right)\)
x(2x^2-x-6)
x(2x^2-4x+3x-6)
x[2x(x-2)+3(x-2)]
x(2x+3)(x-2)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(^{x^3-6x^2-x+30=x^3-5x^2-3x^2+15x-2x^2-10x-6x+30}\)
=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)
\(x^3-6x^2-x+30\)
= \(x^3-5x^2-3x^2+15x+2x^2-10x-6x+30\)
= \(x^2\left(x-5\right)-3x\left(x-5\right)+2x\left(x-5\right)-6\left(x-5\right)\)
= \(\left(x-5\right)\left(x^2-3x+2x-6\right)\)
= \(\left(x-5\right)\left(x\left(x-3\right)+2\left(x-3\right)\right)\)
= \(\left(x-5\right)\left(x+2\right)\left(x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
x3-6x2-x+30
=x3-5x2-x2+5x-6x+30
=(x-5)(x2-x-6)
=(x-5)(x-3)(x+2)
\(x^3-6x^2-x+30=x^3-3x^2-3x^2+9x-10x+30.\)
\(=x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-3x-10\right)\)
\(=\left(x-3\right)\left(x^2+2x-5x-10\right)\)
\(=\left(x-3\right)\left[x\left(x+2\right)-5\left(x+2\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)
Vậy \(x^3-6x^2-x+30=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)