Question

find the product for each of the following expressions. collect all like terms
a) $(9x - 4)(9x + 4)$
b) $(2y - 11)^2$
c) $(5t^3 + 6)^2$
d) $(9z^2 - 1)(9z^2 + 1)$

Explanation:

Part a

Step1: Apply difference of squares rule

Use $(a-b)(a+b)=a^2-b^2$, where $a=9x$, $b=4$.

$$\begin{align*} (9x-4)(9x+4)&=(9x)^2 - 4^2 \end{align*}$$

Step2: Compute each squared term

Calculate the squares of the terms.

$$\begin{align*} (9x)^2 - 4^2&=81x^2 - 16 \end{align*}$$

Part b

Step1: Apply square of binomial rule

Use $(a-b)^2=a^2-2ab+b^2$, where $a=2y$, $b=11$.

$$\begin{align*} (2y-11)^2&=(2y)^2 - 2(2y)(11) + 11^2 \end{align*}$$

Step2: Compute each term

Calculate the squares and product.

$$\begin{align*} (2y)^2 - 2(2y)(11) + 11^2&=4y^2 - 44y + 121 \end{align*}$$

Part c

Step1: Apply square of binomial rule

Use $(a+b)^2=a^2+2ab+b^2$, where $a=5t^3$, $b=6$.

$$\begin{align*} (5t^3+6)^2&=(5t^3)^2 + 2(5t^3)(6) + 6^2 \end{align*}$$

Step2: Compute each term

Calculate the squares and product.

$$\begin{align*} (5t^3)^2 + 2(5t^3)(6) + 6^2&=25t^6 + 60t^3 + 36 \end{align*}$$

Part d

Step1: Apply difference of squares rule

Use $(a-b)(a+b)=a^2-b^2$, where $a=9z^2$, $b=1$.

$$\begin{align*} (9z^2-1)(9z^2+1)&=(9z^2)^2 - 1^2 \end{align*}$$

Step2: Compute each squared term

Calculate the squares of the terms.

$$\begin{align*} (9z^2)^2 - 1^2&=81z^4 - 1 \end{align*}$$

Answer:

a) $81x^2 - 16$
b) $4y^2 - 44y + 121$
c) $25t^6 + 60t^3 + 36$
d) $81z^4 - 1$