Question

两道中二数学题(用十字相乘法)1.(3p+2)(p-3)+4p2.5(3w-1)^2+29(3w-1)-6

Answer 1

1.(3p+2)(p-3)+4p

=3p^2-7p-6+4p

=3p^2-3p-6

=3(p^2-p-2)

=3(p-2)(p+1)

2.5(3w-1)^2+29(3w-1)-6

=[5(3w-1)-1][(3w-1)+6]

=(15w-5-1)(2w-1+6)

=(15w-6)(2w+5)

=3(5w-2)(2w+5)