fumefluosault7pa

2022-02-11

How do you multiply $\left(2{p}^{3}+5{p}^{2}-4\right)\left(3p+1\right)$?

misseymayhemipv

Explanation:
$\left(2{p}^{3}+5{p}^{2}-4\right)\left(3p+1\right)=2{p}^{3}×3p+5{p}^{2}×3p-4×3p+2{p}^{3}+5{p}^{2}-4$
$=6{p}^{4}+15{p}^{3}-12p+2{p}^{3}+5{p}^{2}-4$
$\left(2{p}^{3}+5{p}^{2}-4\right)\left(3p+1\right)=6{p}^{4}+17{p}^{3}+5{p}^{2}-12p-4$

Keenan Mora

By multiplying each part of the right hand bracket, 3p and 1, by each part of the left hand bracket, $2{p}^{3},5{p}^{2}$ and -4 you get:
$6{p}^{4}+15{p}^{3}-12p+2{p}^{3}+5{p}^{2}-4$ which simplifies to:
$6{p}^{4}+17{p}^{3}+5{p}^{2}-12p-4$

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