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


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.

Answer and Explanation:

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{!}{3.2.5!} = 56 \\ \displaystyle \binom{8}{4}=\dfrac{8!}{4! (8-4)!} = \dfrac{!}{!} = 70 \\ \displaystyle \binom{8}{5}=\dfrac{8!}{5! (8-5)!} = \dfrac{!}{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}


{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{!}{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}

Learn more about this topic:

How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16

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