a)\(x^2+2x-8\)
\(\Leftrightarrow x^2+4x-2x-8\)
\(\Leftrightarrow\left(x^2+4x\right)-\left(2x+8\right)\)
\(\Leftrightarrow x\left(x+4\right)-2\left(x+4\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x-2\right)\)
b)\(12x^2-13x+3\)
\(\Leftrightarrow12x^2-9x-4x+3\)
\(\Leftrightarrow\left(12x^2-9x\right)-\left(4x-3\right)\)
\(\Leftrightarrow3x\left(4x-3\right)-\left(4x-3\right)\)
\(\Leftrightarrow\left(4x-3\right)\left(3x-1\right)\)
c)\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(\Leftrightarrow\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(\Leftrightarrow\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=t\) ta có:
\(t\left(t+2\right)-24\)
\(\Leftrightarrow t^2+2t-24\)
\(\Leftrightarrow t^2+6t-4t-24\)
\(\Leftrightarrow\left(t^2+6t\right)-\left(4t+24\right)\)
\(\Leftrightarrow t\left(t+6\right)-4\left(t+6\right)\)
\(\Leftrightarrow\left(t+6\right)\left(t-4\right)\)
Thay t=\(x^2+7x+10\) ta có
\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)\)
\(\Leftrightarrow\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
\(\Leftrightarrow\left(x^2+7x+16\right)\left(x^2+6x+x+6\right)\)
\(\Leftrightarrow\left(x^2+7x+16\right)\left[\left(x^2+6x\right)+\left(x+6\right)\right]\)
\(\Leftrightarrow\left(x^2+7x+16\right)\left[x\left(x+6\right)+\left(x+6\right)\right]\)
\(\Leftrightarrow\left(x^2+7x+16\right)\left(x+6\right)\left(x+1\right)\)
