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GCSEFurther MathsAQARevision Notes2. Algebra & FunctionsBinomial ExpansionProblem Solving with Binomial Expansion

Problem Solving with Binomial Expansion (AQA GCSE Further Maths)

Revision Note

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Problem Solving with Binomial Expansion

How do I find a specific term in a binomial expansion?

  • If asked to find a specific term, use the fact that each term in the expansion of open parentheses a plus b close parentheses to the power of n has the form

    • Pascal space coefficient space cross times a to the power of open parentheses... close parentheses end exponent cross times b to the power of open parentheses... close parentheses end exponent where

      • the Pascal coefficient comes from the row in Pascal's triangle starting with 1 comma space space n comma space space...

      • the powers of a and b sum to the power of the binomial, n

pascals-triangle-no-labels

  • For example, to find the coefficient of the x squared term in the expansion of open parentheses 2 x plus 3 close parentheses to the power of 4

    • Find the row in Pascal's triangle that starts with 1, 4, ...

      • 1, 4, 6, 4, 1

    • Imagine where the power x squared would be in the expansion

Pascal coefficient

1

4

6

4

1

Power of open parentheses 2 x close parentheses

open parentheses 2 x close parentheses to the power of 4

open parentheses 2 x close parentheses cubed

open parentheses 2 x close parentheses squared

open parentheses 2 x close parentheses to the power of 1

open parentheses 2 x close parentheses to the power of 0

Power of open parentheses 3 close parentheses

open parentheses 3 close parentheses to the power of 0

open parentheses 3 close parentheses to the power of 1

open parentheses 3 close parentheses squared

open parentheses 3 close parentheses cubed

open parentheses 3 close parentheses to the power of 4

  • The x squared term must be formed from the middle column, 6 cross times open parentheses 2 x close parentheses squared cross times 3 squared

    • meaning the coefficient of the x squared term is 6 cross times 2 squared cross times 3 squared equals 216

How do I expand binomials with fractions?

  • Some binomials have fractional terms

    • Remember the index law open parentheses a over b close parentheses to the power of n equals a to the power of n over b to the power of n space end exponent

  • For example, open parentheses x plus 1 over x close parentheses to the power of 4 equals x to the power of 4 plus 4 x cubed open parentheses 1 over x close parentheses plus 6 x squared open parentheses 1 over x close parentheses squared plus 4 x open parentheses 1 over x close parentheses cubed plus open parentheses 1 over x close parentheses to the power of 4

    • This simplifies to x to the power of 4 plus 4 x cubed cross times 1 over x plus 6 x squared cross times 1 over x squared plus 4 x cross times 1 over x cubed plus 1 over x to the power of 4

    • Powers of x can then be cancelled

      • x to the power of 4 plus 4 x squared plus 6 plus 4 over x squared plus 1 over x to the power of 4

  • Note how the constant term is no longer at the end of the expansion

Examiner Tips and Tricks

  • Look out for extra information about unknowns

    • e.g. if it says "...where p greater than 0" and you have p equals plus-or-minus 3 then use the positive value

  • If you forget how to find a specific term in the exam, just expand the whole binomial using Pascal's triangle then find it

Worked Example

(a) Find the coefficient of x cubed in the expansion of open parentheses 4 x minus 2 close parentheses to the power of 5.
 

Imagine a equals open parentheses 4 x close parentheses and b equals open parentheses negative 2 close parentheses in open parentheses a plus b close parentheses to the power of 5

The row from Pascal's triangle that starts 1, 5, ... is

table row 1 5 10 10 5 1 end table

The term required is x cubed, which is third along when considering x to the power of 5 comma space x to the power of 4 comma space x cubed comma space...

table row 1 5 cell circle enclose 10 end cell 10 5 1 end table

Therefore the term required is 10 a cubed b squared in the expansion of open parentheses a plus b close parentheses to the power of 5

10 cross times open parentheses 4 x close parentheses cubed cross times open parentheses negative 2 close parentheses squared

So the coefficient of x cubed is

table row cell 10 cross times 4 cubed cross times open parentheses negative 2 close parentheses squared end cell equals cell 10 cross times 64 cross times 4 end cell row blank equals 2560 end table

The coefficient of the x cubed term is 2560

(b) Given that p greater than 0 and that the coefficient of x to the power of 4 in the expansion of open parentheses 3 x minus p close parentheses to the power of 6  is 59 535, find the value of p.

Imagine a equals open parentheses 3 x close parentheses and b equals open parentheses negative p close parentheses in open parentheses a plus b close parentheses to the power of 6

The row from Pascal's triangle that starts 1, 6, ... is

table row 1 6 15 20 15 cell 6 space space space 1 end cell end table

The term required is x to the power of 4, which is third along when considering x to the power of 6 comma space x to the power of 5 comma space x to the power of 4 comma space...

table row 1 6 cell circle enclose 15 end cell 20 15 cell 6 space space space 1 end cell end table

Therefore the term required is 15 a to the power of 4 b squared in the expansion of open parentheses a plus b close parentheses to the power of 6

15 cross times open parentheses 3 x close parentheses to the power of 4 cross times open parentheses negative p close parentheses squared

So the coefficient of x to the power of 4 is

table row cell 15 cross times 3 to the power of 4 cross times open parentheses negative p close parentheses squared end cell equals cell 1215 p squared end cell end table

The question gives the coefficient as 59 535 so set up and solve an equation for p

table row cell 1215 p squared end cell equals cell 59 space 535 end cell row cell p squared end cell equals 49 end table

p greater than 0 so the positive square root is needed

The value of p is 7

(c) Find the coefficient of x in the expansion of open parentheses 2 x plus 1 over x close parentheses to the power of 5.

Imagine a equals open parentheses 2 x close parentheses and b equals open parentheses 1 over x close parentheses in open parentheses a plus b close parentheses to the power of 5

The row from Pascal's triangle that starts 1, 5, ... is

table row 1 5 10 10 5 1 end table

The term required is in x, but this time both terms in the binomial depend on x so consider how their powers are multiplied together in the expansion, open parentheses 2 x close parentheses to the power of 5 open parentheses 1 over x close parentheses to the power of 0 comma space open parentheses 2 x close parentheses to the power of 4 open parentheses 1 over x close parentheses to the power of 1 comma space open parentheses 2 x close parentheses cubed open parentheses 1 over x close parentheses squared comma space...

The third along givesspace 8 x cubed cross times 1 over x squared equals 8 x, so select the third number from the row in Pascal's triangle

table row 1 5 cell circle enclose 10 end cell 10 5 1 end table

The term required is 10 a cubed b squared in the expansion of open parentheses a plus b close parentheses to the power of 5

10 cross times open parentheses 2 x close parentheses cubed cross times open parentheses 1 over x close parentheses squared

So the coefficient of x is

table row cell 10 cross times 2 cubed end cell equals 80 end table

The coefficient of the x term is 80

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    Author
    Reviewer
    Jamie Wood

    Author: Jamie Wood

    Expertise: Maths

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.