How do you multiply #(2p^{2} + p - 3) ( 3p - 1)#?

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Answer 1 · 2016-09-23T06:49:59

#(2p^2+p-3)(3p-1)=6p^3+p^2-10p+3#

To multiply #(2p^2+p-3)(3p-1)#, we use distributive property. According to distributive property,

#axx(b+c)=axxb+axxc#

Hence #(2p^2+p-3)(3p-1)#

= #(2p^2+p-3)xx3p-(2p^2+p-3)xx1# -----------------(1)

and using commutative property of multiplication #axxb=bxxa#

Hence (1) is equal to

#3pxx(2p^2+p-3)-1xx(2p^2+p-3)#

= #3pxx2p^2+3pxxp-3pxx3-2p^2-p+3#

= #6p^3+3p^2-9p-2p^2-p+3#

= #6p^3+p^2-10p+3#