Algebra worksheet for practicing multiplication of monomials by binomials.

Problem: Simplify each expression by multiplying a monomial by a binomial.

The general approach to solving these problems is to use the distributive property. The distributive property states that for any expressions \(a\), \(b\), and \(c\):

\[
a(b + c) = ab + ac
\]

We will apply this property to each problem step by step.

---

Problem 1: \(-3a^3(-8a^4 - 3a^2)\)

1. Distribute \(-3a^3\) to both terms inside the parentheses:
\[
-3a^3(-8a^4) + (-3a^3)(-3a^2)
\]

2. Multiply the coefficients and add the exponents of \(a\):
- For the first term: \((-3)(-8) = 24\) and \(a^3 \cdot a^4 = a^{3+4} = a^7\). So, \(-3a^3(-8a^4) = 24a^7\).
- For the second term: \((-3)(-3) = 9\) and \(a^3 \cdot a^2 = a^{3+2} = a^5\). So, \((-3a^3)(-3a^2) = 9a^5\).

3. Combine the results:
\[
24a^7 + 9a^5
\]

Answer:
\[
\boxed{24a^7 + 9a^5}
\]

---

Problem 2: \(9a^3(-8a^4 + 2a^3)\)

1. Distribute \(9a^3\) to both terms inside the parentheses:
\[
9a^3(-8a^4) + 9a^3(2a^3)
\]

2. Multiply the coefficients and add the exponents of \(a\):
- For the first term: \((9)(-8) = -72\) and \(a^3 \cdot a^4 = a^{3+4} = a^7\). So, \(9a^3(-8a^4) = -72a^7\).
- For the second term: \((9)(2) = 18\) and \(a^3 \cdot a^3 = a^{3+3} = a^6\). So, \(9a^3(2a^3) = 18a^6\).

3. Combine the results:
\[
-72a^7 + 18a^6
\]

Answer:
\[
\boxed{-72a^7 + 18a^6}
\]

---

Problem 3: \(9h^4(-9h - 9)\)

1. Distribute \(9h^4\) to both terms inside the parentheses:
\[
9h^4(-9h) + 9h^4(-9)
\]

2. Multiply the coefficients and add the exponents of \(h\):
- For the first term: \((9)(-9) = -81\) and \(h^4 \cdot h = h^{4+1} = h^5\). So, \(9h^4(-9h) = -81h^5\).
- For the second term: \((9)(-9) = -81\) and \(h^4 \cdot 1 = h^4\). So, \(9h^4(-9) = -81h^4\).

3. Combine the results:
\[
-81h^5 - 81h^4
\]

Answer:
\[
\boxed{-81h^5 - 81h^4}
\]

---

Problem 4: \(4b(9b^5 + 7b^4)\)

1. Distribute \(4b\) to both terms inside the parentheses:
\[
4b(9b^5) + 4b(7b^4)
\]

2. Multiply the coefficients and add the exponents of \(b\):
- For the first term: \((4)(9) = 36\) and \(b \cdot b^5 = b^{1+5} = b^6\). So, \(4b(9b^5) = 36b^6\).
- For the second term: \((4)(7) = 28\) and \(b \cdot b^4 = b^{1+4} = b^5\). So, \(4b(7b^4) = 28b^5\).

3. Combine the results:
\[
36b^6 + 28b^5
\]

Answer:
\[
\boxed{36b^6 + 28b^5}
\]

---

Problem 5: \(-8v^3(3v^5 + 3v^4)\)

1. Distribute \(-8v^3\) to both terms inside the parentheses:
\[
-8v^3(3v^5) + (-8v^3)(3v^4)
\]

2. Multiply the coefficients and add the exponents of \(v\):
- For the first term: \((-8)(3) = -24\) and \(v^3 \cdot v^5 = v^{3+5} = v^8\). So, \(-8v^3(3v^5) = -24v^8\).
- For the second term: \((-8)(3) = -24\) and \(v^3 \cdot v^4 = v^{3+4} = v^7\). So, \((-8v^3)(3v^4) = -24v^7\).

3. Combine the results:
\[
-24v^8 - 24v^7
\]

Answer:
\[
\boxed{-24v^8 - 24v^7}
\]

---

Problem 6: \(-2a^2(-8a^2 + 9a)\)

1. Distribute \(-2a^2\) to both terms inside the parentheses:
\[
-2a^2(-8a^2) + (-2a^2)(9a)
\]

2. Multiply the coefficients and add the exponents of \(a\):
- For the first term: \((-2)(-8) = 16\) and \(a^2 \cdot a^2 = a^{2+2} = a^4\). So, \(-2a^2(-8a^2) = 16a^4\).
- For the second term: \((-2)(9) = -18\) and \(a^2 \cdot a = a^{2+1} = a^3\). So, \((-2a^2)(9a) = -18a^3\).

3. Combine the results:
\[
16a^4 - 18a^3
\]

Answer:
\[
\boxed{16a^4 - 18a^3}
\]

---

Problem 7: \(7a^2(-3a^4 + 6a^3)\)

1. Distribute \(7a^2\) to both terms inside the parentheses:
\[
7a^2(-3a^4) + 7a^2(6a^3)
\]

2. Multiply the coefficients and add the exponents of \(a\):
- For the first term: \((7)(-3) = -21\) and \(a^2 \cdot a^4 = a^{2+4} = a^6\). So, \(7a^2(-3a^4) = -21a^6\).
- For the second term: \((7)(6) = 42\) and \(a^2 \cdot a^3 = a^{2+3} = a^5\). So, \(7a^2(6a^3) = 42a^5\).

3. Combine the results:
\[
-21a^6 + 42a^5
\]

Answer:
\[
\boxed{-21a^6 + 42a^5}
\]

---

Problem 8: \(-6p(2p^3 - 4p^2)\)

1. Distribute \(-6p\) to both terms inside the parentheses:
\[
-6p(2p^3) + (-6p)(-4p^2)
\]

2. Multiply the coefficients and add the exponents of \(p\):
- For the first term: \((-6)(2) = -12\) and \(p \cdot p^3 = p^{1+3} = p^4\). So, \(-6p(2p^3) = -12p^4\).
- For the second term: \((-6)(-4) = 24\) and \(p \cdot p^2 = p^{1+2} = p^3\). So, \((-6p)(-4p^2) = 24p^3\).

3. Combine the results:
\[
-12p^4 + 24p^3
\]

Answer:
\[
\boxed{-12p^4 + 24p^3}
\]

---

Problem 9: \(-2p^3(-4p^4 - 7p^3)\)

1. Distribute \(-2p^3\) to both terms inside the parentheses:
\[
-2p^3(-4p^4) + (-2p^3)(-7p^3)
\]

2. Multiply the coefficients and add the exponents of \(p\):
- For the first term: \((-2)(-4) = 8\) and \(p^3 \cdot p^4 = p^{3+4} = p^7\). So, \(-2p^3(-4p^4) = 8p^7\).
- For the second term: \((-2)(-7) = 14\) and \(p^3 \cdot p^3 = p^{3+3} = p^6\). So, \((-2p^3)(-7p^3) = 14p^6\).

3. Combine the results:
\[
8p^7 + 14p^6
\]

Answer:
\[
\boxed{8p^7 + 14p^6}
\]

---

Problem 10: \(3x^3(-2x^5 + 9x^4)\)

1. Distribute \(3x^3\) to both terms inside the parentheses:
\[
3x^3(-2x^5) + 3x^3(9x^4)
\]

2. Multiply the coefficients and add the exponents of \(x\):
- For the first term: \((3)(-2) = -6\) and \(x^3 \cdot x^5 = x^{3+5} = x^8\). So, \(3x^3(-2x^5) = -6x^8\).
- For the second term: \((3)(9) = 27\) and \(x^3 \cdot x^4 = x^{3+4} = x^7\). So, \(3x^3(9x^4) = 27x^7\).

3. Combine the results:
\[
-6x^8 + 27x^7
\]

Answer:
\[
\boxed{-6x^8 + 27x^7}
\]

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Final Answers:

1. \(\boxed{24a^7 + 9a^5}\)
2. \(\boxed{-72a^7 + 18a^6}\)
3. \(\boxed{-81h^5 - 81h^4}\)
4. \(\boxed{36b^6 + 28b^5}\)
5. \(\boxed{-24v^8 - 24v^7}\)
6. \(\boxed{16a^4 - 18a^3}\)
7. \(\boxed{-21a^6 + 42a^5}\)
8. \(\boxed{-12p^4 + 24p^3}\)
9. \(\boxed{8p^7 + 14p^6}\)
10. \(\boxed{-6x^8 + 27x^7}\)