[Đại 8] Phân tích đa thức thành nhân tử

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tiasangmangtenss

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cherrynguyen_298

1
.$x^{3}+2x-3 $
=$x^2(x-1)+x(x-1)+3(x-1)$
=$(x-1)(x^2+x+3)$

 
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thienbinhgirl

1, $x^3-x^2+x^2-x+3x-3=x^2(x-1)+x(x-1)+3(x-1)=(x-1)(x^2+x+3)$......................................................................
 
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cherrynguyen_298

2.
$x^{3}+5x^{2}+8x+4$
=$x^{3}+x^{2}+4x^{2}+4x+4x+4$
=$x^{2} ( x+1 )+4x ( x+1 )+4 ( x+1 )$
=$ (x+1) ( x+2 )^{2}$
 
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thienbinhgirl

3..........................................$x^3-2x^2+x^2-2x+x-2$
$=x^2(x-2)+x(x-2)+(x-2)$
$=(x^2+x+1)(x-2)$
 
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thienbinhgirl

4.........................$x^3+2x^2-x^2+2x-x+2$
$=x^2(x+2)-x(x+2)+(x+2)$
$=(x^2-x+1)(x+2)$
 
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cherrynguyen_298

7.
$x^{3}+5x^{2}+3x-9$
=$x^{3}-x^{2}+6x^{2}-6x+9x-9$
=$x^{2} ( x-1 )+6x ( x-1 )+9 ( x-1 )$
=$( x-1 ) ( x+3 )^{2}$
 
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thienbinhgirl

6.....................................
$x^3-x^2+6x^2-6x+9x-9$
$=x^2(x-1)+6x(x-1)+9(x-1)$
$=(x-1)(x+3)^2$
 
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thienbinhgirl

8......................................
$x^3+2x^2+7x^2+14x+12x+24$
$=x^2(x+2)+7x(x+2)+12(x+2)$
$=(x+2)(x+6)(x+1)$
 
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sapanang

$6)x^3+9x^2+11x−21$

$6)x^3+9x^2+11x−21$
$=x^3+7x^2+3x^2+21x-x^2-7x-3x-21$
$=X^2(x+7)+3x(x+7)-x(x+7)-3(x+7)$
$=(x^2+3x-x-3)(x+7)$
$=[x(x+3)-(x+3)](x+7)$
=(x-1)(x+3)(x+7)
 
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luongmanhkhoa

1)x^3+2x−3
2)x^3+5x^2+8x+4
3)x^3−x^2−x−2
4)x^3+x^2−x+2
5)x^3−5x^2−x^2+5x−6x+30
6)x^3+9x^2+11x−21
7)x^3+5x^2+3x−9
8)x^3+9x^2+26x+24
-------------------------------
1/ x^3+2x-3
= x^2 + 2x+1 -4x
= x^2-2x+1
= (x+1)^2
2/ x^3+5x^2+8x+4
=x^3+x^2+4x^2+4x+4x+4
=x^2(x+1)+4x(x+1)+4(x+1)
=(x+1)(x+2)^2
3/ x^3-x^2-x-2
=x^2(x−2)+x(x−2)+(x−2)
=(x2+x+1)(x−2)
4/ x^3+2x^2−x^2+2x−x+2
=x^2(x+2)−x(x+2)+(x+2)
=(x2−x+1)(x+2)
5/ ---------------------------
=x^2(x−2)−x(x−5)−6(x−5)
=(x+2)(x−3)(x−5)
6)x^3+9x^2+11x−21
=x^3−x^2+6x^2−6x+9x−9
=x^2(x−1)+6x(x−1)+9(x−1)
=(x−1)(x+3)^2
7)x^3+5x^2+3x−9
=x^3−x^2+6x^2−6x+9x−9
=x^2(x−1)+6x(x−1)+9(x−1)
=(x−1)(x+3)2
 
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