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

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

#### Explanation:

$\left(2 {p}^{3} + 5 {p}^{2} - 4\right) \left(3 p + 1\right) = 2 {p}^{3} \times 3 p + 5 {p}^{2} \times 3 p - 4 \times 3 p + 2 {p}^{3} + 5 {p}^{2} - 4$

$= 6 {p}^{4} + 15 {p}^{3} - 12 p + 2 {p}^{3} + 5 {p}^{2} - 4$

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

Mar 3, 2018

$6 {p}^{4} + 17 {p}^{3} + 5 {p}^{2} - 12 p - 4$

#### Explanation:

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

$6 {p}^{4} + 15 {p}^{3} - 12 p + 2 {p}^{3} + 5 {p}^{2} - 4$ which simplifies to:

$6 {p}^{4} + 17 {p}^{3} + 5 {p}^{2} - 12 p - 4$

Mar 3, 2018

$6 {p}^{4} + 17 {p}^{3} + 5 {p}^{2} - 12 p - 4$

#### 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:
$= \left(2 {p}^{3} \cdot 3 p\right) + \left(2 {p}^{3} \cdot 1\right) + \left(5 {p}^{2} \cdot 3 p\right) + \left(5 {p}^{2} \cdot 1\right) + \left(- 4 \cdot 3 p\right) + \left(- 4 \cdot 1\right)$
This simplifies to:
$= 6 {p}^{4} + 2 {p}^{3} + 15 {p}^{3} + 5 {p}^{2} - 12 p - 4$
$= 6 {p}^{4} + 17 {p}^{3} + 5 {p}^{2} - 12 p - 4$

Note: When multiplying variables with the same base you add their exponents
${p}^{3} = p \cdot p \cdot p$ therefore ${p}^{3} \cdot p = p \cdot p \cdot p \cdot p = {p}^{4}$