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Find the coefficient of x^8 in expanding (x-2)^{11}.

Question:

Find the coefficient of {eq}x^8 {/eq} in expanding {eq}(x-2)^{11} {/eq}.

Binomial expansion

The expansion of the polynomial equation to the nth term so as to know the different terms at different locations. It is the algebraic sum of the binomial expansion of power. The binomial expansion is for the polynomial equation only.

Answer and Explanation:


The expression in the form of equation of the binomial equation is given as

{eq}{\left( {x - 2} \right)^{11}}...........................(I) {/eq}


Consider the above equation as

{eq}\begin{align*} a &= \left( { - 2} \right)\\ b &= \left( x \right) \end{align*} {/eq}


The expansion of {eq}{\left( {a + b} \right)^n} {/eq} is

{eq}{\left( {a + b} \right)^n} = {a^n}{ + ^n}{C_1}{a^{n - 1}}b + ......... + {b^n} {/eq}


Expand the equation (I) in form of the above equation

The expression for the term in the GP series is

{eq}{P_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r} {/eq}


Here the{eq}n {/eq} is the total number of term.

To find the coefficient of the{eq}{x^8} {/eq}, the value of the ratio {eq}r = 8 {/eq}


Substitute the value in the above equation

{eq}\begin{align*} &{P_{8 + 1}}{= ^{11}}{C_8}{\left( { - 2} \right)^{11 - 8}}{\left( x \right)^8}\\ &= \left( {\dfrac{{11!}}{{\left( {11 - 8} \right)! \times 8!}}} \right)\left( { - 8} \right){x^8}\\ &= - 1320{x^8} \end{align*} {/eq}


Thus the value of the coefficient is {eq}- 1320 {/eq}


Learn more about this topic:

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How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
9.8K

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