1. Section 3.1
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| 2x | 5 | |
| x | 2x^2 | 5x |
| 6 | 12x | 30 |
We get ( 2x+5)( x+6)=2x^2+5x+12x+30=2x^2+17x+30.
| m | -3 | |
| 3m | 3m^2 | -9m |
| 5 | 5m | -15 |
That way, we get ( m-3)( 3m+5)=3m^2-9m+5m-15=3m^2-4m-15.
| 12x | 1 | |
| x^2 | 12x^3 | x^2 |
| -5 | -60x | -5 |
Hence, ( 12x+1)( x^2-5)=12x^3+x^2-60x-5.
| 3 | -5y | |
| 2 | 6 | -10y |
| y | 3y | -5y^2 |
Consequently, ( 3-5y)( 2+y)=6-10y+3y-5y^2=-5y^2-7y+6.