Class work exercise on expanding binomial products with step-by-step examples.

To solve the problem, we need to expand each product of binomials. The general method for expanding \((x + a)(x + b)\) is:

\[
(x + a)(x + b) = x^2 + (a + b)x + ab
\]

Let's go through each expression step by step.

---

1. \((x + 3)(x + 5)\)

Using the formula:
\[
(x + 3)(x + 5) = x^2 + (3 + 5)x + (3 \cdot 5)
\]
\[
= x^2 + 8x + 15
\]

---

2. \((x + 4)(x + 6)\)

Using the formula:
\[
(x + 4)(x + 6) = x^2 + (4 + 6)x + (4 \cdot 6)
\]
\[
= x^2 + 10x + 24
\]

---

3. \((x + 3)(x + 7)\)

Using the formula:
\[
(x + 3)(x + 7) = x^2 + (3 + 7)x + (3 \cdot 7)
\]
\[
= x^2 + 10x + 21
\]

---

4. \((x + 5)(x + 8)\)

Using the formula:
\[
(x + 5)(x + 8) = x^2 + (5 + 8)x + (5 \cdot 8)
\]
\[
= x^2 + 13x + 40
\]

---

5. \((x + 6)(x + 8)\)

Using the formula:
\[
(x + 6)(x + 8) = x^2 + (6 + 8)x + (6 \cdot 8)
\]
\[
= x^2 + 14x + 48
\]

---

6. \((x + 4)(x + 12)\)

Using the formula:
\[
(x + 4)(x + 12) = x^2 + (4 + 12)x + (4 \cdot 12)
\]
\[
= x^2 + 16x + 48
\]

---

7. \((x + 3)(x + 16)\)

Using the formula:
\[
(x + 3)(x + 16) = x^2 + (3 + 16)x + (3 \cdot 16)
\]
\[
= x^2 + 19x + 48
\]

---

8. \((x + 2)(x + 24)\)

Using the formula:
\[
(x + 2)(x + 24) = x^2 + (2 + 24)x + (2 \cdot 24)
\]
\[
= x^2 + 26x + 48
\]

---

9. \((x + 1)(x + 48)\)

Using the formula:
\[
(x + 1)(x + 48) = x^2 + (1 + 48)x + (1 \cdot 48)
\]
\[
= x^2 + 49x + 48
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&(x + 3)(x + 5) = x^2 + 8x + 15 \\
&(x + 4)(x + 6) = x^2 + 10x + 24 \\
&(x + 3)(x + 7) = x^2 + 10x + 21 \\
&(x + 5)(x + 8) = x^2 + 13x + 40 \\
&(x + 6)(x + 8) = x^2 + 14x + 48 \\
&(x + 4)(x + 12) = x^2 + 16x + 48 \\
&(x + 3)(x + 16) = x^2 + 19x + 48 \\
&(x + 2)(x + 24) = x^2 + 26x + 48 \\
&(x + 1)(x + 48) = x^2 + 49x + 48
\end{aligned}
}
\]