Questions: Find each product. A. (6n+4p)^2 B. (4m^3-2l)^2 C. (2b^2-g)(2b^2+g)

Transcript text: 22. Find each product. A. $(6 n+4 p)^{2}$ B. $\left(4 m^{3}-2 l\right)^{2}$ C. $\left(2 b^{2}-g\right)\left(2 b^{2}+g\right)$

Solution

Solution Steps

To solve these problems, we will use algebraic identities to expand the expressions. For part A and B, we will use the square of a binomial formula: $(a + b)^2 = a^2 + 2ab + b^2$. For part C, we will use the difference of squares formula: $(a - b)(a + b) = a^2 - b^2$.

Step 1: Expand $ (6n + 4p)^2 $

Using the formula $ (a + b)^2 = a^2 + 2ab + b^2 $, we have: \[ (6n + 4p)^2 = (6n)^2 + 2(6n)(4p) + (4p)^2 = 36n^2 + 48np + 16p^2 \] Substituting $ n = 1 $ and $ p = 1 $: \[ 36(1)^2 + 48(1)(1) + 16(1)^2 = 36 + 48 + 16 = 100 \]

Step 2: Expand $ (4m^3 - 2l)^2 $

Using the same formula, we have: \[ (4m^3 - 2l)^2 = (4m^3)^2 + 2(4m^3)(-2l) + (-2l)^2 = 16m^6 - 16m^3l + 4l^2 \] Substituting $ m = 1 $ and $ l = 1 $: \[ 16(1)^6 - 16(1)^3(1) + 4(1)^2 = 16 - 16 + 4 = 4 \]

Step 3: Expand $ (2b^2 - g)(2b^2 + g) $

Using the difference of squares formula, we have: \[ (2b^2 - g)(2b^2 + g) = (2b^2)^2 - g^2 = 4b^4 - g^2 \] Substituting $ b = 1 $ and $ g = 1 $: \[ 4(1)^4 - (1)^2 = 4 - 1 = 3 \]