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

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

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - (x - 1)

= (x³ - 1)(x - 1)

x^5+x^4+1

=x⁵+x⁴+x³+x²+x+1-x³-x²-x

=x³.(x²+x+1)+(x²+x+1)-x.(x²+x+1)

tự xử tiếp

Minh Triều?????

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

Ta có: \(x^{m+4}-x^{m+3}-x+1\)

\(=x^{m+3}\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(x^{m+3}-1\right)\)

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

\(x^4-2x^3+2x-1\)

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

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

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

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

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

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

\(x^4+2x^3+x^2+x+1\)

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

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

\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

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

\(=(x+4-1)(x+4+1)(x-1)(x+1)\)

\(=(x+3)(x+5)(x-1)(x+1)\)

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

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

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

\(=\left(x^2-1\right)\left(x+3\right)\left(x+5\right)\)

Cái này đã là nhân tử rồi mà bạn

\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

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

\(=\)\(x^2+8x-9\)

\(=x^2+9x-x-9\)

\(=(x-1)(x+9)\)

\(5x^2-4\left(x^2-2x+1\right)-5\)

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

\(=x^2+8x-9\)

\(=\left(x+9\right)\left(x-1\right)\)

Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)

\(=4\left(x^2+60+17x\right)\left(x^2+60x+16x\right)+3x^2\)

\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)

\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)