# Write out the binomial expansion for each binomial raised to the 8th power. (Be sure to show all...

## Question:

Write out the binomial expansion for each binomial raised to the 8th power. (Be sure to show all steps and evaluate the binomial coefficients and write in simplified form.)

A. {eq}(x - y)^8 {/eq}

B. {eq}(2a + 3b)^8 {/eq}

Without writing out the whole polynomial, find the 16th term for {eq}(x + y)^{19} {/eq}. List the steps of how you reached your answer.

## Binomial Expansion or Newton's Binomial:

The expansion of a binomial has some properties that should be known by the student, among them we can mention the following: when the terms have different signs, we have to alternate positive and negative signs in the development of the binomial. There are n+1 terms for a power n.

The binomial expansion of a binomial raised to {eq}nth {/eq} power is given by the formula

{eq}{\color{Red}{ (a+b)^n = \displaystyle \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k }} {/eq}

A) For {eq}(x-y)^8 {/eq} we observe that the sign of {eq}y {/eq} is negative and therefore the odd degree powers will be negative. In this case, we use the formula

{eq}{\color{Red}{ (a-b)^n = \displaystyle \sum_{k=0}^n(-1)^k \binom{n}{k} a^{n-k}b^k }} {/eq}

This way, we have

{eq}\begin{align*} (x-y)^8 &= \displaystyle \sum_{k=0}^8(-1)^k \binom{8}{k} x^{8-k}y^k \\ &= (-1)^0\binom{8}{0}x^{8-0}y^0 + (-1)^1\binom{8}{1}x^{8-1}y^1 +(-1)^2\binom{8}{2}x^{8-2}y^2 + (-1)^3\binom{8}{3}x^{8-3}y^3 + (-1)^4\binom{8}{4}x^{8-4}y^4 + \\ & + (-1)^5\binom{8}{5}x^{8-5}y^5 +(-1)^6\binom{8}{6}x^{8-6}y^6 + (-1)^7\binom{8}{7}x^{8-7}y^7 +(-1)^8\binom{8}{8}x^{8-8}y^8 \\ &= \binom{8}{0}x^{8} -\binom{8}{1}x^{7}y +\binom{8}{2}x^{6}y^2 -\binom{8}{3}x^{5}y^3 + \binom{8}{4}x^{4}y^4 -\binom{8}{5}x^{3}y^5 +\binom{8}{6}x^{2}y^6 -\binom{8}{7}xy^7 +\binom{8}{8}y^8 \end{align*} {/eq}

Remember that the development of the binomial coefficient is:

{eq}\displaystyle \binom{n}{k}=\dfrac{n!}{k!(n-k)!} {/eq}, thus

{eq}\displaystyle \binom{8}{0}=\dfrac{8!}{0!(8-0)!} = \dfrac{8!}{1.8!} = 1 \\ \displaystyle \binom{8}{1}=\dfrac{8!}{1! (8-1)!} = \dfrac{8.7!}{7!} =8 \\ \displaystyle \binom{8}{2}=\dfrac{8!}{2! (8-2)!} = \dfrac{8.7.6!}{2.6!} = 28 \\ \displaystyle \binom{8}{3}=\dfrac{8!}{3! (8-3)!} = \dfrac{8.7.6.5!}{3.2.5!} = 56 \\ \displaystyle \binom{8}{4}=\dfrac{8!}{4! (8-4)!} = \dfrac{8.7.6.5.4!}{4.3.2.4!} = 70 \\ \displaystyle \binom{8}{5}=\dfrac{8!}{5! (8-5)!} = \dfrac{8.7.6.5!}{5!3.2} = 56 \\ \displaystyle \binom{8}{6}=\dfrac{8!}{6! (8-6)!} = \dfrac{8.7.6!}{6!2} = 28 \\ \displaystyle \binom{8}{7}=\dfrac{8!}{7! (8-7)!} = \dfrac{8.7!}{7!1} =8 \\ \displaystyle \binom{8}{8}=\dfrac{8!}{8!(8-8)!} = \dfrac{8!}{8!1} = 1 \\ {/eq}

Remember {eq}0!=1 {/eq} and {eq}1!=1 {/eq}

Finally

{eq}(x-y)^8= x^{8} -8x^{7}y +28x^{6}y^2 - 56x^{5}y^3 + 70x^{4}y^4 -56x^{3}y^5 +28x^{2}y^6 -8xy^7 +y^8 {/eq}

B) For {eq}(2a+3)^8 {/eq} We use the binomial coefficients calculated in part A and the formula for positive binomial terms

{eq}(2a+3)^8= 1(2a)^{8} + 8(2a)^{7}3 +28(2a)^{6}3^2 + 56(2a)^{5}3^3 + 70(2a)^{4}3^4 + 56(2a)^{3}3^5 +28(2a)^{2}3^6 + 8(2a)3^7 +3^8 {/eq}

Perform the powers and multiplying numbers we have

{eq}(2a+3)^8=256a^8+3072a^7+16128a^6+48384a^5+90720a^4+108864a^3+81648a^2+34992a+6561 \\ {/eq}

Find the 16th term for {eq}(x + y)^{19} {/eq}

Step 1: We use the formula

{eq}{\color{Red}{ (x+y)^n = \displaystyle \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k }} {/eq}

With {eq}n=19 {/eq} and{eq}k=16 {/eq}

Step 2: The formula for any term is given by

{eq}t_k=\binom{n}{k-1}x^{n-(k-1)}y^{k-1} \\ {/eq}

Step 3: We substitute the data

{eq}t_{16}=\binom{19}{16-1}x^{19-(16-1)}y^{16-1} \\ {/eq}

Step 4: We perform the calculations

{eq}\displaystyle \binom{19}{15}= \dfrac{19!}{15!(19-15)!}= \dfrac{19.18.17.16.15!}{15!4!}=3876 \\ x^{19-(16-1)}=x^4 \\ y^{16-1}=y^{15} {/eq}

Step 5: We write the term 16th

{eq}t_{16}=3876x^4y^{15} {/eq}

How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
10K