# How do you multiply polynomials [4 - (3c - 1)][6 - ( 3c - 1)]?

Oct 15, 2015

$9 {c}^{2} - 36 c + 35$

#### Explanation:

First of all, you can simplify both expressions inside the square bracket: for the first one, you get

$4 - \left(3 c - 1\right) = 4 - 3 c - \left(- 1\right) = 4 - 3 c + 1 = - 3 c + 5$

In the same fashion, for the second square bracket you get

6−(3c−1)=-3c+7

Your multiplication is now written as

$\left(- 3 c + 5\right) \left(- 3 c + 7\right)$. To do such a multiplication, you need to do all the possible multiplications of the terms in the first bracket with those of the second, and them sum them up:

1. First term times first term: $\left(- 3 c\right) \cdot \left(- 3 c\right) = 9 {c}^{2}$;
2. First term times second term: $\left(- 3 c\right) \cdot 7 = - 21 c$;
3. Second term times first term: $5 \cdot \left(- 3 c\right) = - 15 c$
4. **Second term times second term: $5 \cdot 7 = 35$.

Now we sum them up: $9 {c}^{2} - 21 c - 15 c + 35$ equals

$9 {c}^{2} - 36 c + 35$