The Factor Theorem (Edexcel IGCSE Maths B): Revision Note

Exam code: 4MB1

Roger B

Written by: Roger B

Reviewed by: Jamie Wood

Updated on

Factor Theorem

What is the factor theorem?

  • The factor theorem is used to find the linear factors of a function

    • This is closely related to finding the roots (or solutions) of a function or equation

  • For a function straight f open parentheses x close parentheses, the factor theorem tells us that

    • If  straight f open parentheses a close parentheses equals 0, then open parentheses x minus a close parentheses is a factor of straight f open parentheses x close parentheses

    •  If open parentheses x minus a close parentheses is a factor of straight f open parentheses x close parentheses, then  straight f open parentheses a close parentheses equals 0

How do I use the factor theorem?

  • Consider the  function straight f open parentheses x close parentheses where open parentheses x minus a close parentheses is a factor

    • Then by the factor theorem we know that straight f open parentheses a close parentheses equals 0

      • I.e., x equals a is a solution to the equation  straight f open parentheses x close parentheses equals 0

  • Or consider the function straight f open parentheses x close parentheses where straight f open parentheses a close parentheses equals 0

    • Then by the factor theorem we know  that open parentheses x minus a close parentheses is a factor of straight f open parentheses x close parentheses

    • Therefore  straight f left parenthesis x right parenthesis equals left parenthesis x minus a right parenthesis cross times Q left parenthesis x right parenthesis

      • where Q open parentheses x close parentheses is a function that is also a factor of straight f open parentheses x close parentheses

    • Hence  fraction numerator straight f left parenthesis x right parenthesis over denominator open parentheses x minus a close parentheses end fraction equals Q left parenthesis x right parenthesis

      • I.e. Q open parentheses x close parentheses is the quotient when straight f open parentheses x close parentheses is divided by open parentheses x minus a close parentheses

      • And the remainder is equal to zero

  • If the linear factor has a coefficient of x (other than 1) you must first factorise out the coefficient

    • For the linear factor  left parenthesis b x blank – blank c right parenthesis blank equals b open parentheses x minus c over b close parentheses

      • straight f open parentheses c over b close parentheses equals 0

      • straight f open parentheses x close parentheses equals b open parentheses x minus c over b close parentheses cross times Q open parentheses x close parentheses equals open parentheses b x minus c close parentheses cross times Q open parentheses x close parentheses

Examiner Tips and Tricks

Be careful with the minus sign in a factor open parentheses x minus a close parentheses.

  • That means a is a solution to f open parentheses x close parentheses equals 0, not negative a !

If you are looking for integer solutions to straight f open parentheses x close parentheses equals 0  (where straight f open parentheses x close parentheses is a polynomial)

  • those solutions will always be factors of the constant term in straight f open parentheses x close parentheses

Worked Example

(a) Consider the functionspace straight f left parenthesis x right parenthesis equals x cubed minus 2 x squared minus x plus 2. Given that x equals 2 is a solution to the equation straight f open parentheses x close parentheses equals 0, write down a linear factor of straight f open parentheses x close parentheses.

Answer:

By the factor theorem, if  straight f open parentheses a close parentheses equals 0 then open parentheses x minus a close parentheses is a factor of straight f open parentheses x close parentheses

x minus 2 

(b) Use the factor theorem to determine whether open parentheses x plus 1 close parentheses is a factor of space straight g left parenthesis x right parenthesis equals 2 x cubed plus 3 x squared minus x plus 5.

Answer:

By the factor theorem, open parentheses x minus a close parentheses can only be a factor of straight g open parentheses x close parentheses if  straight g open parentheses a close parentheses equals 0

But be careful, here a is equal to negative 1, not 1

table row cell straight g open parentheses negative 1 close parentheses end cell equals cell 2 open parentheses negative 1 close parentheses cubed plus 3 open parentheses negative 1 close parentheses squared minus open parentheses negative 1 close parentheses plus 5 end cell row blank equals cell negative 2 plus 3 plus 1 plus 5 end cell row blank equals 7 end table

straight g stretchy left parenthesis negative 1 stretchy right parenthesis not equal to 0, so Error converting from MathML to accessible text. is not a factor of straight g stretchy left parenthesis x stretchy right parenthesis

(c) Given that open parentheses 2 x minus 3 close parentheses is a factor of space straight h open parentheses x close parentheses equals 2 x cubed minus b x squared plus 7 x minus 6, find the value of b.

Answer:

open parentheses 2 x minus 3 close parentheses equals 2 open parentheses x minus 3 over 2 close parentheses,  so open parentheses x minus 3 over 2 close parentheses is a factor of straight h open parentheses x close parentheses

Therefore by the factor theorem, straight h open parentheses 3 over 2 close parentheses equals 0

table row cell space 2 open parentheses 3 over 2 close parentheses cubed minus b open parentheses 3 over 2 close parentheses squared plus 7 open parentheses 3 over 2 close parentheses minus 6 end cell equals 0 row cell 27 over 4 minus 9 over 4 b plus 21 over 2 minus 6 end cell equals 0 row cell 45 over 4 minus 9 over 4 b end cell equals 0 row cell 9 over 4 b end cell equals cell 45 over 4 end cell row b equals cell 4 over 9 cross times 45 over 4 end cell end table

b equals 5

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Roger B

Author: Roger B

Expertise: Maths Content Creator

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Jamie Wood

Reviewer: Jamie Wood

Expertise: Maths Content Creator

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.