Bài 2:
a, \(x^3+21x^2+134x+240\)
\(=x^3+3x^2+18x^2+54x+80x+240\)
\(=x^2\left(x+3\right)+18x\left(x+3\right)+8\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+18x+8\right)\)
\(=\left(x+3\right)\left(x^2+8x+10x+80\right)\)
\(=\left(x+3\right)\left(x+8\right)\left(x+10\right)\)
b, \(x^5-x^4-1\)
\(=x^5-x^3-x^2-x^4+x^2+x+x^3-x-1\)
\(=x^2\left(x^3-x-1\right)-x\left(x^3-x-1\right)+\left(x^3-x-1\right)\)
\(=\left(x^3-x-1\right)\left(x^2-x+1\right)\)
Chúc bạn học tốt!!!
a, \(f\left(x\right)=\left(x^2+x\right)^2+4x^2+4x-12\)
\(f\left(x\right)=x^4+2x^3+x^2+4x^2+4x-12\)
\(f\left(x\right)=x^4+2x^3+5x^2+4x-12\)
\(f\left(x\right)=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(f\left(x\right)=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(f\left(x\right)=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(f\left(x\right)=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(f\left(x\right)=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(f\left(x\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
b, \(g\left(x\right)=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(g\left(x\right)=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(g\left(x\right)=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+10\Rightarrow t+2=x^2+7x+12\)
Khi đó \(g\left(x\right)=t.\left(t+2\right)-24=t^2+2t-24\)
\(g\left(x\right)=t^2-4t+6t-24\)
\(g\left(x\right)=t\left(t-4\right)+6\left(t-4\right)=\left(t-4\right)\left(t+6\right)\)
Vì \(t=x^2+7x+10\) nên
\(g\left(x\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(g\left(x\right)=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(g\left(x\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Chúc bạn học tốt!!!