# How do you find the product (3b^2-4b-7)(2b^2-b-9)?

Nov 7, 2017

$6 {b}^{4} - 11 {b}^{3} - 34 {b}^{2} + 43 b + 63$

#### Explanation:

This will be extremely tedious, though not difficult. Please take very close care of your negatives and exponents as you follow along.

We take the first term of the first polynomial, and multiply it by each term of the second polynomial. We then repeat this process with the second and third term of the first polynomial. Here we go.

First term
$3 {b}^{2} \cdot 2 {b}^{2} = 6 {b}^{4}$
$3 {b}^{2} \cdot - b = - 3 {b}^{3}$
$3 {b}^{2} \cdot - 9 = - 27 {b}^{2}$

Second term
$- 4 b \cdot 2 {b}^{2} = - 8 {b}^{3}$
$- 4 b \cdot - b = 4 {b}^{2}$
$- 4 b \cdot - 9 = 36 b$

Third term
$- 7 \cdot 2 {b}^{2} = - 14 {b}^{2}$
$- 7 \cdot - b = 7 b$
$- 7 \cdot - 9 = 63$

Now, we take the answer to all 9 of our problems and put them in one long equation.

$6 {b}^{4} \textcolor{red}{- 3 {b}^{3}} \textcolor{b l u e}{- 27 {b}^{2}} \textcolor{red}{- 8 {b}^{3}} \textcolor{b l u e}{+ 4 {b}^{2}} \textcolor{g r e e n}{+ 36 b} \textcolor{b l u e}{- 14 {b}^{2}} \textcolor{g r e e n}{+ 7 b} + 63$

Note there are a few terms you can combine here to further simplify. These are color coded. Add them together to simplify and get your final answer:
$6 {b}^{4} \textcolor{red}{- 11 {b}^{3}} \textcolor{b l u e}{- 34 {b}^{2}} \textcolor{g r e e n}{+ 43 b} + 63$