# How do you find the product (3c+4d)^2?

Jan 24, 2017

${\left(3 c + 4 d\right)}^{2} = 9 {c}^{2} + 24 c d + 16 {d}^{2}$

#### Explanation:

${\left(3 c + 4 d\right)}^{2}$

= $\left(3 c + 4 d\right) \times \left(3 c + 4 d\right)$

= $3 c \times \left(3 c + 4 d\right) + 4 d \times \left(3 c + 4 d\right)$

= $3 c \times 3 c + 3 c \times 4 d + 4 d \times 3 c + 4 d \times 4 d$

= $9 {c}^{2} + 12 c d + 12 c d + 16 {d}^{2}$

= $9 {c}^{2} + 24 c d + 16 {d}^{2}$

Jan 24, 2017

${\left(3 c + 4 d\right)}^{2} = 9 {c}^{2} + 24 c d + 16 {d}^{2}$

#### Explanation:

Using the special product ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$, we can set $a = 3 c$ and $b = 4 d$ to get

${\left(3 c + 4 d\right)}^{2} = {\left(3 c\right)}^{2} + 2 \left(3 c\right) \left(4 d\right) + {\left(4 d\right)}^{2}$

$= {3}^{2} {c}^{2} + \left(2\right) \left(3\right) \left(4\right) c d + {4}^{2} {d}^{2}$

$= 9 {c}^{2} + 24 c d + 16 {d}^{2}$