How do you multiply #(2p^3+5p^2-4)(3p+1)#?

3 Answers
Shiva Prakash M V
Mar 3, 2018

#(2p^3+5p^2-4)(3p+1)=6p^4+17p^3+5p^2-12p-4#

Explanation:

#(2p^3+5p^2-4)(3p+1)=2p^3xx3p+5p^2xx3p-4xx3p+2p^3+5p^2-4#

#=6p^4+15p^3-12p+2p^3+5p^2-4#

#(2p^3+5p^2-4)(3p+1)=6p^4+17p^3+5p^2-12p-4#

OliKing
Mar 3, 2018

#6p^4+17p^3+5p^2-12p-4#

Explanation:

By multiplying each part of the right hand bracket, #3p# and #1#, by each part of the left hand bracket, #2p^3#, #5p^2# and #-4# you get:

#6p^4+15p^3-12p+2p^3+5p^2-4# which simplifies to:

#6p^4+17p^3+5p^2-12p-4#

Aphiwe Shange
Mar 3, 2018

#6p^4+17p^3+5p^2-12p-4#

Explanation:

you take the sum of products of each term in the first bracket with each term in the second. so with this example its would be:
#=(2p^3 * 3p)+(2p^3 * 1)+(5p^2 * 3p)+(5p^2 * 1)+(-4 * 3p)+(-4 * 1)#
This simplifies to:
#=6p^4+2p^3+15p^3+5p^2-12p-4#
#=6p^4+17p^3+5p^2-12p-4#

Note: When multiplying variables with the same base you add their exponents
#p^3=p*p*p# therefore #p^3*p=p*p*p*p=p^4#

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