Answer key for a printable math worksheet on multiplying binomials, featuring eight problems with solutions in red text.

It looks like you've uploaded an answer key for a worksheet titled *"Multiplying Binomials"*, which includes 8 problems involving the multiplication of two binomials. The correct answers are already provided in red.

However, since you're asking to solve the problem and explain the solution, I'll go through each problem step by step, showing how to multiply the binomials using the FOIL method (First, Outer, Inner, Last) or distributive property, and verify that the results match the given answers.

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🔷 Problem 1:

$$
(-9p + 11)(6p - 3)
$$

Step-by-step:

Use FOIL:

- F: $(-9p)(6p) = -54p^2$
- O: $(-9p)(-3) = 27p$
- I: $(11)(6p) = 66p$
- L: $(11)(-3) = -33$

Now combine:
$$
-54p^2 + 27p + 66p - 33 = -54p^2 + 93p - 33
$$

✔ Matches the answer key.

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🔷 Problem 2:

$$
(-7q^6 - q^4)(-2q^2 - q^5)
$$

Distribute each term:

First, distribute $-7q^6$:
- $-7q^6 \cdot (-2q^2) = 14q^8$
- $-7q^6 \cdot (-q^5) = 7q^{11}$

Then distribute $-q^4$:
- $-q^4 \cdot (-2q^2) = 2q^6$
- $-q^4 \cdot (-q^5) = q^9$

Now collect all terms:
$$
7q^{11} + q^9 + 14q^8 + 2q^6
$$

✔ Matches the answer key.

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🔷 Problem 3:

$$
(-5y^4 - y^5)(-8y^5 + 2y^6)
$$

Distribute:

First, $-5y^4$:
- $-5y^4 \cdot (-8y^5) = 40y^9$
- $-5y^4 \cdot 2y^6 = -10y^{10}$

Then, $-y^5$:
- $-y^5 \cdot (-8y^5) = 8y^{10}$
- $-y^5 \cdot 2y^6 = -2y^{11}$

Now combine:
$$
-2y^{11} + (-10y^{10} + 8y^{10}) + 40y^9 = -2y^{11} - 2y^{10} + 40y^9
$$

✔ Matches the answer key.

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🔷 Problem 4:

$$
\left(\frac{7}{8}b - \frac{3}{4}\right)\left(\frac{6}{7}b^3 + 24\right)
$$

Distribute:

First, $\frac{7}{8}b$:
- $\frac{7}{8}b \cdot \frac{6}{7}b^3 = \frac{42}{56}b^4 = \frac{3}{4}b^4$
- $\frac{7}{8}b \cdot 24 = \frac{168}{8}b = 21b$

Then, $-\frac{3}{4}$:
- $-\frac{3}{4} \cdot \frac{6}{7}b^3 = -\frac{18}{28}b^3 = -\frac{9}{14}b^3$
- $-\frac{3}{4} \cdot 24 = -18$

Combine:
$$
\frac{3}{4}b^4 - \frac{9}{14}b^3 + 21b - 18
$$

✔ Matches the answer key.

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🔷 Problem 5:

$$
(11w^6 + w^3)(5w^3 + 1)
$$

Distribute:

$11w^6$:
- $11w^6 \cdot 5w^3 = 55w^9$
- $11w^6 \cdot 1 = 11w^6$

$w^3$:
- $w^3 \cdot 5w^3 = 5w^6$
- $w^3 \cdot 1 = w^3$

Now combine:
$$
55w^9 + (11w^6 + 5w^6) + w^3 = 55w^9 + 16w^6 + w^3
$$

✔ Matches the answer key.

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🔷 Problem 6:

$$
(-2c + c^2)(-4c + c^2)
$$

Rewrite in standard form: $(c^2 - 2c)(c^2 - 4c)$

Distribute:

$c^2 \cdot c^2 = c^4$
$c^2 \cdot (-4c) = -4c^3$
$-2c \cdot c^2 = -2c^3$
$-2c \cdot (-4c) = 8c^2$

Now combine:
$$
c^4 + (-4c^3 - 2c^3) + 8c^2 = c^4 - 6c^3 + 8c^2
$$

✔ Matches the answer key.

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🔷 Problem 7:

$$
\left(\frac{4}{5}a^5 - 4a\right)\left(-\frac{1}{3}a^6 - \frac{5}{3}a^2\right)
$$

Distribute:

First, $\frac{4}{5}a^5$:
- $\frac{4}{5}a^5 \cdot (-\frac{1}{3}a^6) = -\frac{4}{15}a^{11}$
- $\frac{4}{5}a^5 \cdot (-\frac{5}{3}a^2) = -\frac{20}{15}a^7 = -\frac{4}{3}a^7$

Then, $-4a$:
- $-4a \cdot (-\frac{1}{3}a^6) = \frac{4}{3}a^7$
- $-4a \cdot (-\frac{5}{3}a^2) = \frac{20}{3}a^3$

Now combine:
- $-\frac{4}{15}a^{11}$
- $-\frac{4}{3}a^7 + \frac{4}{3}a^7 = 0$
- $\frac{20}{3}a^3$

So final result:
$$
-\frac{4}{15}a^{11} + \frac{20}{3}a^3
$$

✔ Matches the answer key.

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🔷 Problem 8:

$$
(6x + 9)(6x - 9)
$$

This is a difference of squares pattern:
$$
(a + b)(a - b) = a^2 - b^2
$$

Here, $a = 6x$, $b = 9$:
$$
(6x)^2 - (9)^2 = 36x^2 - 81
$$

✔ Matches the answer key.

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✔ Final Summary:

All 8 problems have been solved correctly using distribution (FOIL), combining like terms, and simplifying fractions where needed. Each result matches the provided answer key.

Let me know if you'd like a printable version or a PDF explanation!