Question

6.3
module 6.3: exit ticket

1. use the binomial theorem to expand each power of a binomial.

(2x - 1)³

1. use the binomial theorem to expand each power of a binomial.

(4x + 3y)⁶

1. use the binomial theorem to find the specified term of the given power of a binomial. (remember that r starts at 0 in the binomial theorem, so finding, say, the second term means that r = 1.)

find the third term in the expanded form of (3x - 2y)⁵.

Explanation:

Problem 1: Expand \((2x - 1)^3\) 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 \(n!=n\times(n - 1)\times\cdots\times1\) (with \(0!=1\)). For \((2x-1)^3\), we have \(a = 2x\), \(b=- 1\), and \(n = 3\).

Step 2: Calculate each term for \(k = 0,1,2,3\)

• For \(k = 0\):

\(\binom{3}{0}(2x)^{3-0}(-1)^{0}=\frac{3!}{0!3!}(8x^{3})(1)=1\times8x^{3}\times1 = 8x^{3}\)

• For \(k = 1\):

\(\binom{3}{1}(2x)^{3 - 1}(-1)^{1}=\frac{3!}{1!2!}(4x^{2})(-1)=\frac{3\times2!}{1\times2!}\times4x^{2}\times(-1)=3\times4x^{2}\times(-1)=-12x^{2}\)

• For \(k = 2\):

\(\binom{3}{2}(2x)^{3-2}(-1)^{2}=\frac{3!}{2!1!}(2x)(1)=\frac{3\times2!}{2!\times1}\times2x\times1 = 3\times2x\times1=6x\)

• For \(k = 3\):

\(\binom{3}{3}(2x)^{3-3}(-1)^{3}=\frac{3!}{3!0!}(1)(-1)=1\times1\times(-1)=-1\)

Step 3: Sum the terms

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

Step 1: Recall the Binomial Theorem

For \((4x + 3y)^6\), \(a = 4x\), \(b = 3y\), \(n=6\). The Binomial Theorem is \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\)

Step 2: Calculate each term for \(k=0\) to \(k = 6\)

• \(k = 0\): \(\binom{6}{0}(4x)^{6}(3y)^{0}=\frac{6!}{0!6!}(4096x^{6})(1)=1\times4096x^{6}\times1 = 4096x^{6}\)
• \(k = 1\): \(\binom{6}{1}(4x)^{5}(3y)^{1}=\frac{6!}{1!5!}(1024x^{5})(3y)=\frac{6\times5!}{1\times5!}\times1024x^{5}\times3y=6\times1024x^{5}\times3y = 18432x^{5}y\)
• \(k = 2\): \(\binom{6}{2}(4x)^{4}(3y)^{2}=\frac{6!}{2!4!}(256x^{4})(9y^{2})=\frac{6\times5\times4!}{2\times1\times4!}\times256x^{4}\times9y^{2}=15\times256x^{4}\times9y^{2}=34560x^{4}y^{2}\)
• \(k = 3\): \(\binom{6}{3}(4x)^{3}(3y)^{3}=\frac{6!}{3!3!}(64x^{3})(27y^{3})=\frac{6\times5\times4\times3!}{3!\times3\times2\times1}\times64x^{3}\times27y^{3}=20\times64x^{3}\times27y^{3}=34560x^{3}y^{3}\)
• \(k = 4\): \(\binom{6}{4}(4x)^{2}(3y)^{4}=\frac{6!}{4!2!}(16x^{2})(81y^{4})=\frac{6\times5\times4!}{4!\times2\times1}\times16x^{2}\times81y^{4}=15\times16x^{2}\times81y^{4}=19440x^{2}y^{4}\)
• \(k = 5\): \(\binom{6}{5}(4x)^{1}(3y)^{5}=\frac{6!}{5!1!}(4x)(243y^{5})=\frac{6\times5!}{5!\times1}\times4x\times243y^{5}=6\times4x\times243y^{5}=5832xy^{5}\)
• \(k = 6\): \(\binom{6}{6}(4x)^{0}(3y)^{6}=\frac{6!}{6!0!}(1)(729y^{6})=1\times1\times729y^{6}=729y^{6}\)

Step 3: Sum the terms

\((4x + 3y)^6=4096x^{6}+18432x^{5}y + 34560x^{4}y^{2}+34560x^{3}y^{3}+19440x^{2}y^{4}+5832xy^{5}+729y^{6}\)

Step 1: Recall the Binomial Theorem and term numbering

The Binomial Theorem is \((a + b)^n=\sum_{r = 0}^{n}\binom{n}{r}a^{n - r}b^{r}\), where the term corresponding to \(r\) is the \((r + 1)\)-th term. We need the third term, so \(r+1 = 3\) which means \(r = 2\). Here, \(a = 3x\), \(b=-2y\), and \(n = 5\).

Step 2: Calculate the term for \(r = 2\)

First, calculate \(\binom{5}{2}=\frac{5!}{2!3!}=\frac{5\times4\times3!}{2\times1\times3!}=10\)

Then, \(a^{n - r}=(3x)^{5 - 2}=(3x)^{3}=27x^{3}\)

And \(b^{r}=(-2y)^{2}=4y^{2}\)

Now, multiply these together with \(\binom{5}{2}\):

\(\binom{5}{2}(3x)^{3}(-2y)^{2}=10\times27x^{3}\times4y^{2}=10\times108x^{3}y^{2}=1080x^{3}y^{2}\)

Answer:

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

Problem 2: Expand \((4x + 3y)^6\) using the Binomial Theorem