\(\left\{{}\begin{matrix}a+b=m\\a-b=n\end{matrix}\right.\)\(\Rightarrow4ab=m^2-n^2\)

Ta có:

\(A=\left(m+c\right)^3-4.\dfrac{m^3+3mn^2}{4}-4c^3-3c\left(m^2-n^2\right)\)

\(=3.\left(-c^3+mc^2-mn^2+cn^2\right)\)

\(=3.\left(m-c\right).\left(c+n\right).\left(c-n\right)\)

\(\Rightarrow A=3.\left(a+b-c\right).\left(c+a-b\right).\left(c-a+b\right)\)

phân tích đa thức thành nhân tử

4x^4+4x^3-x^2-x

x^6-x^4-9x^3+9x^2

x^4-4x^3+8x^2-16x+16

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

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

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

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

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

\(=x^2\left(x^4-x^2\right)-x^2\left(9x-9\right)\)

\(=x^2\left(x^4-x^2-9x+9\right)\)

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

\(=x^2\left[x\left(x^3+x^2-9\right)-\left(x^3+x^2-9\right)\right]\)

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

\(x^4-4x^3+8x^2-16x+16\)

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

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

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

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

Phân tích đa thức đa thức

a. \(x^3-5x^2+8x-4\)

b. \(x^3-3x+2\)

c. \(x^3-5x^2+3x+9\)

d. \(x^3+8x^2+17x+10\)

e. \(x^3+3x^2+6x+4\)

f. \(x^3+3x^2+3x+2\)

******************************************************

a) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4\)

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

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

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

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

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

c) \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9\)

\(=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-6x+9\right)=\left(x+1\right)\left(x-3\right)^2\)

d) \(x^3+8x^2+17x+10=x^3+2x^2+6x^2+12x+5x+10\)

\(=x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+6x+5\right)=\left(x+2\right)\left(x+5\right)\left(x+1\right)\)

e) \(x^3+3x^2+6x+4=x^3+x^2+2x^2+2x+4x+4\)

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

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

f) \(x^3+3x^2+3x+2=x^3+2x^2+x^2+2x+x+2\)

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

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