Question

name teacher hour for 16–17, use the binomial theorem to expand each power of a binomial. 16. $(x - 2)^5$ 17. $(2x + 1)^3$ original content copyright © by houghton mifflin harcourt. additions and changes to the original content are the responsibility of the instructor. algebra 2b unit 4 test

Explanation:

Problem 16: Expand \((x - 2)^5\) using the Binomial Theorem

Step 1: Recall the Binomial Theorem

The Binomial Theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n!=n\times(n - 1)\times\cdots\times1\). For \((x-2)^5\), we have \(a = x\), \(b=-2\), and \(n = 5\).

Step 2: Calculate each term

• For \(k = 0\): \(\binom{5}{0}x^{5-0}(-2)^{0}=\frac{5!}{0!5!}x^{5}\times1 = 1\times x^{5}\times1=x^{5}\)
• For \(k = 1\): \(\binom{5}{1}x^{5 - 1}(-2)^{1}=\frac{5!}{1!4!}x^{4}\times(-2)=\frac{5\times4!}{1\times4!}x^{4}\times(-2)=5\times(-2)x^{4}=- 10x^{4}\)
• For \(k = 2\): \(\binom{5}{2}x^{5-2}(-2)^{2}=\frac{5!}{2!3!}x^{3}\times4=\frac{5\times4\times3!}{2\times1\times3!}x^{3}\times4 = 10\times4x^{3}=40x^{3}\)
• For \(k = 3\): \(\binom{5}{3}x^{5-3}(-2)^{3}=\frac{5!}{3!2!}x^{2}\times(-8)=\frac{5\times4\times3!}{3!\times2\times1}x^{2}\times(-8)=10\times(-8)x^{2}=-80x^{2}\)
• For \(k = 4\): \(\binom{5}{4}x^{5-4}(-2)^{4}=\frac{5!}{4!1!}x^{1}\times16=\frac{5\times4!}{4!\times1}x\times16 = 5\times16x=80x\)
• For \(k = 5\): \(\binom{5}{5}x^{5-5}(-2)^{5}=\frac{5!}{5!0!}x^{0}\times(-32)=1\times1\times(-32)=-32\)

Step 3: Sum up all the terms

\((x - 2)^5=x^{5}-10x^{4}+40x^{3}-80x^{2}+80x - 32\)

Step 1: Recall the Binomial Theorem

The Binomial Theorem is \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). For \((2x + 1)^3\), we have \(a = 2x\), \(b = 1\), and \(n=3\).

Step 2: Calculate each term

• For \(k = 0\): \(\binom{3}{0}(2x)^{3-0}(1)^{0}=\frac{3!}{0!3!}(8x^{3})\times1=1\times8x^{3}\times1 = 8x^{3}\)
• For \(k = 1\): \(\binom{3}{1}(2x)^{3 - 1}(1)^{1}=\frac{3!}{1!2!}(4x^{2})\times1=\frac{3\times2!}{1\times2!}\times4x^{2}=3\times4x^{2}=12x^{2}\)
• For \(k = 2\): \(\binom{3}{2}(2x)^{3-2}(1)^{2}=\frac{3!}{2!1!}(2x)\times1=\frac{3\times2!}{2!\times1}\times2x=3\times2x = 6x\)
• For \(k = 3\): \(\binom{3}{3}(2x)^{3-3}(1)^{3}=\frac{3!}{3!0!}(1)\times1=1\times1\times1 = 1\)

Step 3: Sum up all the terms

\((2x + 1)^3=8x^{3}+12x^{2}+6x + 1\)

Answer:

\(x^{5}-10x^{4}+40x^{3}-80x^{2}+80x - 32\)

Problem 17: Expand \((2x + 1)^3\) using the Binomial Theorem