A)\(x^2+5x-6=x^2-x+6x-6\)
\(=x\left(x-1\right)+6\left(x-1\right)\)
\(=\left(x+6\right)\left(x-1\right)\)
x + 6 = 0 x = - 6
| x - 1 = 0 x = 1 |
b, \(2x^2+3x-5=2\left(x^2+\frac{3}{2}x-\frac{5}{2}\right)\)
\(=2\left(x^2-x+\frac{5}{2}x-\frac{5}{2}\right)\)
\(=2\left[x\left(x-1\right)+\frac{5}{2}\left(x-1\right)\right]\)
\(=2\left(x+\frac{5}{2}\right)\left(x-1\right)\)
\(\left(x+\frac{5}{2}\right)=0\) \(x=\frac{-5}{2}\) | \(\left(x-1\right)=0\) \(x=1\) |
