[Đại 8] Phân tích đa thức thành nhân tử

K

khaiproqn81

Like gấp !!

$x^3-7x-6=x^3+x^2-x^2-x-6x-6 \\ =x^2(x+1)-x(x+1)-6(x+1) = (x+1)(x^2-x-6) \\ =(x+1)(x^2+2x-3x-6)=(x+1)[x(x+2)-3(x+2)] \\ =(x+1)(x+2)(x-3)$
 
M

minhhieu2468

toán

$x^3-7x-6=x^3+x^2-x^2-x-6x-6=x^2(x+1)-x(x+1)-6(x+1)=(x+1)(X^2-x-6)=0$\Rightarrowx=-1\forallx=3\forallx=-2
 
L

lenhanhieu

Ta có:
x^3-7x-6
= x^3+3x^2+2x-3x^2-9x-6
= (x^3+3x^2+2x)-(3x^2+9x+6)
= x(x^2+3x+2)-3(x^2+3x+2)
= (x-3)(x^2+3x+2)
= (x-3)(x^2+2x+x+2)
= (x-3)(x(x+2)+(x+2))
= (x-3)(x+1)(x+2)

:khi (197):
 
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