\(x^4+3x^3+12x-16\)

\(=x^4+4x^3+4x^2+16x-x^3-4x^2-4x-16\)

\(=x\left(x^3+4x^2+4x+16\right)-\left(x^3+4x^2+4x+16\right)\)

\(=\left(x-1\right)\left(x^3+4x^2+4x+16\right)\)

\(=\left(x-1\right)\left[x^2\left(x+4\right)+4\left(x+4\right)\right]\)

\(=\left(x-1\right)\left(x+4\right)\left(x^2+4\right)\)

\(=2x^4+6x^3-3x^3-9x^2-3x^2-9x+2x+6\)

\(=2x^3\left(x+3\right)-3x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)

\(=\left(x+3\right)\left(2x^3-4x^2+x^2-2x-x+2\right)=\left(x+3\right)\left(x-2\right)\left(2x^2+x-1\right)\)

\(=\left(x+3\right)\left(x-2\right)\left(2x^2+2x-x-1\right)=\left(x+3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)\)

2x^4+3x^3-12x^2-7x+6 = (2x^4-x^3)+(4x^3-2x^2)-(10x^2-5x)-(12x-6)

= x^3.(2x-1)+2x^2.(2x-1)-5x.(2x-1)-6.(2x-1) = (2x-1).(x^3+2x^2-5x-6) 

= (2x-1).[ (x^3+x^2)+(x^2+x)-(6x+6) ] = (2x-1).(x+1).(x^2+x-6) = (2x-1).(x-1).[(x^2-2x)+(3x-6)]

= (2x-1).(x+1).(x-2).(x+3)

k mk nha

\(2x^4+3x^3-12x^2-7x+6\)

\(=\left(2x^4+2x^3\right)+\left(x^3+x^2\right)-\left(13x^2+13x\right)+\left(6x+6\right)\)

\(=\left(x+1\right)\left(2x^3+x^2-13x+6\right)\)

a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

b) \(-x^3-3x^2-3x-1=-\left(x^3+3x^2+3x+1\right)=-\left(x+1\right)^3\)

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

a) \(x^3+3x^2+3x=0\Rightarrow x\left(x^2+3x+3\right)=0\Rightarrow x\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\Rightarrow x=0\)

(do \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

b) \(x^3+6x^2+12x=0\Rightarrow x\left(x^2+6x+12\right)=0\Rightarrow x\left[\left(x+3\right)^2+4\right]=0\Rightarrow x=0\)

(do (x+3)²+4≥4>0)

a: Ta có: \(x^3+3x^2+3x=0\)

\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)

hay x=0

b: Ta có: \(x^3+6x^2+12x=0\)

\(\Leftrightarrow x\left(x^2+6x+12\right)=0\)

hay x=0

\(3x^2-12x=3x.\left(x-12\right)\)

~ hok tốt ~

3x(x-2)

~~~~~~~
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^_^

Phân tích thành nhân tử : 

( Bằng cách đặt nhân tử chung)

\(3x^2-12x\)

\(=3x\left(x-12\right)\)

_Chúc bạn học tốt_

\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)

\(=\left(x^2-x-6\right)^2-x^2=\left(x^2-6\right)\left(x^2-2x-6\right)\)

-3x⁴y-12x²y-12x²y

=-3xy(x^2+4x+4x)

Hk tốt

\(-3x^4y-12x^2y-12x^2y\)

=\(3x^2y\left(-x^2-4-4\right)\)

\(=-3x^2y\left(x^2+8\right)\)

(sai hay đúng thì ko chắc đâu)

\(-3x^4-12x^2y-12xy^2\)

\(=-3x\left(x^2+4xy+4y^2\right)\)

\(=-3x\left(x+2y\right)^2\)

học tốt ~

\(4x^4y-4x^2y^3+12x^3y+12x^2y^2\)

\(=4x^2y\left(x^2-y^2+3x+3y\right)\)

\(=4x^2y\left(x-y-3\right)\left(x+y\right)\)

[(4x+1)(3x+2)][(12x-1)(x+1)]=4

=>(12x^2+11x+2)(12x^2+11x-1)=4

dat 12x^2+11x-1=ythi y(y+3)=4

=>Y^2+3y-4=0

=>y^2+4y-y-4=0

=>y(y+4)-(y+4)=0=>9y-1)(y-4)=0

ban tu giai tiep nha

đa thức này đã phân tích thành nhân tử rồi bạn ạ