Question

1. $(3x + 4)^4$
2. $(4a - 5b)^3$
3. $(5x - 3y)^4$

Explanation:

Problem 14: Expand \((3x + 4)^4\)

We use the binomial theorem: \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n = 4\), \(a=3x\), \(b = 4\).

Step 1: Calculate the binomial coefficients for \(n = 4\)

• For \(k = 0\): \(\binom{4}{0}=\frac{4!}{0!(4 - 0)!}=1\)
• For \(k = 1\): \(\binom{4}{1}=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4\)
• For \(k = 2\): \(\binom{4}{2}=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!\times2!}=6\)
• For \(k = 3\): \(\binom{4}{3}=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=4\)
• For \(k = 4\): \(\binom{4}{4}=\frac{4!}{4!(4 - 4)!}=1\)

Step 2: Expand using the binomial theorem

\[

$$\begin{align*} (3x+4)^{4}&=\binom{4}{0}(3x)^{4}(4)^{0}+\binom{4}{1}(3x)^{3}(4)^{1}+\binom{4}{2}(3x)^{2}(4)^{2}+\binom{4}{3}(3x)^{1}(4)^{3}+\binom{4}{4}(3x)^{0}(4)^{4}\\ &=1\times81x^{4}\times1 + 4\times27x^{3}\times4+6\times9x^{2}\times16 + 4\times3x\times64+1\times1\times256\\ &=81x^{4}+432x^{3}+864x^{2}+768x + 256 \end{align*}$$

\]

Problem 16: Expand \((4a-5b)^{3}\)

We use the binomial theorem: \((a - b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}(-b)^{k}\), where \(n = 3\), \(a = 4a\), \(b=5b\)

Step 1: Calculate the binomial coefficients for \(n = 3\)

• For \(k = 0\): \(\binom{3}{0}=\frac{3!}{0!(3 - 0)!}=1\)
• For \(k = 1\): \(\binom{3}{1}=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3\)
• For \(k = 2\): \(\binom{3}{2}=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=3\)
• For \(k = 3\): \(\binom{3}{3}=\frac{3!}{3!(3 - 3)!}=1\)

Step 2: Expand using the binomial theorem

\[

$$\begin{align*} (4a-5b)^{3}&=\binom{3}{0}(4a)^{3}(-5b)^{0}+\binom{3}{1}(4a)^{2}(-5b)^{1}+\binom{3}{2}(4a)^{1}(-5b)^{2}+\binom{3}{3}(4a)^{0}(-5b)^{3}\\ &=1\times64a^{3}\times1+3\times16a^{2}\times(- 5b)+3\times4a\times25b^{2}+1\times1\times(-125b^{3})\\ &=64a^{3}-240a^{2}b + 300ab^{2}-125b^{3} \end{align*}$$

\]

Problem 18: Expand \((5x-3y)^{4}\)

We use the binomial theorem: \((a - b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}(-b)^{k}\), where \(n = 4\), \(a = 5x\), \(b = 3y\)

Step 1: Calculate the binomial coefficients for \(n = 4\) (same as in problem 14 for \(n = 4\))

\(\binom{4}{0}=1\), \(\binom{4}{1}=4\), \(\binom{4}{2}=6\), \(\binom{4}{3}=4\), \(\binom{4}{4}=1\)

Step 2: Expand using the binomial theorem

\[

$$\begin{align*} (5x-3y)^{4}&=\binom{4}{0}(5x)^{4}(-3y)^{0}+\binom{4}{1}(5x)^{3}(-3y)^{1}+\binom{4}{2}(5x)^{2}(-3y)^{2}+\binom{4}{3}(5x)^{1}(-3y)^{3}+\binom{4}{4}(5x)^{0}(-3y)^{4}\\ &=1\times625x^{4}\times1+4\times125x^{3}\times(-3y)+6\times25x^{2}\times9y^{2}+4\times5x\times(-27y^{3})+1\times1\times81y^{4}\\ &=625x^{4}-1500x^{3}y + 1350x^{2}y^{2}-540xy^{3}+81y^{4} \end{align*}$$

\]

Final Answers

1. \(\boldsymbol{81x^{4}+432x^{3}+864x^{2}+768x + 256}\)

1. \(\boldsymbol{64a^{3}-240a^{2}b + 300ab^{2}-125b^{3}}\)

1. \(\boldsymbol{625x^{4}-1500x^{3}y + 1350x^{2}y^{2}-540xy^{3}+81y^{4}}\)

Answer:

Problem 14: Expand \((3x + 4)^4\)

We use the binomial theorem: \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n = 4\), \(a=3x\), \(b = 4\).

Step 1: Calculate the binomial coefficients for \(n = 4\)

• For \(k = 0\): \(\binom{4}{0}=\frac{4!}{0!(4 - 0)!}=1\)
• For \(k = 1\): \(\binom{4}{1}=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4\)
• For \(k = 2\): \(\binom{4}{2}=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!\times2!}=6\)
• For \(k = 3\): \(\binom{4}{3}=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=4\)
• For \(k = 4\): \(\binom{4}{4}=\frac{4!}{4!(4 - 4)!}=1\)

Step 2: Expand using the binomial theorem

\[

$$\begin{align*} (3x+4)^{4}&=\binom{4}{0}(3x)^{4}(4)^{0}+\binom{4}{1}(3x)^{3}(4)^{1}+\binom{4}{2}(3x)^{2}(4)^{2}+\binom{4}{3}(3x)^{1}(4)^{3}+\binom{4}{4}(3x)^{0}(4)^{4}\\ &=1\times81x^{4}\times1 + 4\times27x^{3}\times4+6\times9x^{2}\times16 + 4\times3x\times64+1\times1\times256\\ &=81x^{4}+432x^{3}+864x^{2}+768x + 256 \end{align*}$$

\]

Problem 16: Expand \((4a-5b)^{3}\)

We use the binomial theorem: \((a - b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}(-b)^{k}\), where \(n = 3\), \(a = 4a\), \(b=5b\)

Step 1: Calculate the binomial coefficients for \(n = 3\)

• For \(k = 0\): \(\binom{3}{0}=\frac{3!}{0!(3 - 0)!}=1\)
• For \(k = 1\): \(\binom{3}{1}=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3\)
• For \(k = 2\): \(\binom{3}{2}=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=3\)
• For \(k = 3\): \(\binom{3}{3}=\frac{3!}{3!(3 - 3)!}=1\)

Step 2: Expand using the binomial theorem

\[

$$\begin{align*} (4a-5b)^{3}&=\binom{3}{0}(4a)^{3}(-5b)^{0}+\binom{3}{1}(4a)^{2}(-5b)^{1}+\binom{3}{2}(4a)^{1}(-5b)^{2}+\binom{3}{3}(4a)^{0}(-5b)^{3}\\ &=1\times64a^{3}\times1+3\times16a^{2}\times(- 5b)+3\times4a\times25b^{2}+1\times1\times(-125b^{3})\\ &=64a^{3}-240a^{2}b + 300ab^{2}-125b^{3} \end{align*}$$

\]

Problem 18: Expand \((5x-3y)^{4}\)

We use the binomial theorem: \((a - b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}(-b)^{k}\), where \(n = 4\), \(a = 5x\), \(b = 3y\)

Step 1: Calculate the binomial coefficients for \(n = 4\) (same as in problem 14 for \(n = 4\))

\(\binom{4}{0}=1\), \(\binom{4}{1}=4\), \(\binom{4}{2}=6\), \(\binom{4}{3}=4\), \(\binom{4}{4}=1\)

Step 2: Expand using the binomial theorem

\[

$$\begin{align*} (5x-3y)^{4}&=\binom{4}{0}(5x)^{4}(-3y)^{0}+\binom{4}{1}(5x)^{3}(-3y)^{1}+\binom{4}{2}(5x)^{2}(-3y)^{2}+\binom{4}{3}(5x)^{1}(-3y)^{3}+\binom{4}{4}(5x)^{0}(-3y)^{4}\\ &=1\times625x^{4}\times1+4\times125x^{3}\times(-3y)+6\times25x^{2}\times9y^{2}+4\times5x\times(-27y^{3})+1\times1\times81y^{4}\\ &=625x^{4}-1500x^{3}y + 1350x^{2}y^{2}-540xy^{3}+81y^{4} \end{align*}$$

\]

Final Answers

1. \(\boldsymbol{81x^{4}+432x^{3}+864x^{2}+768x + 256}\)

1. \(\boldsymbol{64a^{3}-240a^{2}b + 300ab^{2}-125b^{3}}\)

1. \(\boldsymbol{625x^{4}-1500x^{3}y + 1350x^{2}y^{2}-540xy^{3}+81y^{4}}\)