How do you multiply (2p^3+5p^2-4)(3p+1)(2p3+5p2−4)(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(2p3+5p2−4)(3p+1)=6p4+17p3+5p2−12p−4

Explanation:

(2p^3+5p^2-4)(3p+1)=2p^3xx3p+5p^2xx3p-4xx3p+2p^3+5p^2-4(2p3+5p2−4)(3p+1)=2p3×3p+5p2×3p−4×3p+2p3+5p2−4

=6p^4+15p^3-12p+2p^3+5p^2-4=6p4+15p3−12p+2p3+5p2−4

(2p^3+5p^2-4)(3p+1)=6p^4+17p^3+5p^2-12p-4(2p3+5p2−4)(3p+1)=6p4+17p3+5p2−12p−4

OliKing
Mar 3, 2018

6p^4+17p^3+5p^2-12p-46p4+17p3+5p2−12p−4

Explanation:

By multiplying each part of the right hand bracket, 3p3p and 11, by each part of the left hand bracket, 2p^32p3, 5p^25p2 and -4−4 you get:

6p^4+15p^3-12p+2p^3+5p^2-46p4+15p3−12p+2p3+5p2−4 which simplifies to:

6p^4+17p^3+5p^2-12p-46p4+17p3+5p2−12p−4

Aphiwe Shange
Mar 3, 2018

6p^4+17p^3+5p^2-12p-46p4+17p3+5p2−12p−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)=(2p3⋅3p)+(2p3⋅1)+(5p2⋅3p)+(5p2⋅1)+(−4⋅3p)+(−4⋅1)
This simplifies to:
=6p^4+2p^3+15p^3+5p^2-12p-4=6p4+2p3+15p3+5p2−12p−4
=6p^4+17p^3+5p^2-12p-4=6p4+17p3+5p2−12p−4

Note: When multiplying variables with the same base you add their exponents
p^3=p*p*pp3=p⋅p⋅p therefore p^3*p=p*p*p*p=p^4p3⋅p=p⋅p⋅p⋅p=p4

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