# How do you find the product of (2t^2+t+3)(4t^2+2t-2)?

Apr 26, 2017

#### Explanation:

The method for performing the multiplication will become obvious when we write it in a form where each term of the first factor multiplies the second factor:

$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$

$2 {t}^{2} \left(4 {t}^{2} + 2 t - 2\right) +$

$t \left(4 {t}^{2} + 2 t - 2\right) +$

$3 \left(4 {t}^{2} + 2 t - 2\right)$

Distribute the first term of the first factor into the second factor:

$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$

$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$

$t \left(4 {t}^{2} + 2 t - 2\right) +$

$3 \left(4 {t}^{2} + 2 t - 2\right)$

Distribute the second term of the first factor into the second factor:

$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$

$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$

$\textcolor{w h i t e}{\text{.........}} 4 {t}^{3} + 2 {t}^{2} - 2 t +$

$3 \left(4 {t}^{2} + 2 t - 2\right)$

Distribute the third term of the first factor into the second factor:

$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$

$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$

$\textcolor{w h i t e}{\text{.........}} 4 {t}^{3} + 2 {t}^{2} - 2 t +$

$\textcolor{w h i t e}{\text{.................}} 12 {t}^{2} + 6 t - 6$

I have aligned the columns so that they can be added:

$\left(2 {t}^{2} + t + 3\right) \left(4 {t}^{2} + 2 t - 2\right) =$

$8 {t}^{4} + 4 {t}^{3} - 4 {t}^{2} +$

$\textcolor{w h i t e}{\text{.........}} 4 {t}^{3} + 2 {t}^{2} - 2 t +$

$\underline{\textcolor{w h i t e}{\text{.................}} 12 {t}^{2} + 6 t - 6}$
$8 {t}^{4} + 8 {t}^{3} + 10 {t}^{2} + 4 t - 6$

Because multiplication is commutative, we could have multiplied the first factor by each term of the second factor and obtained the same result.