GCSEFurther MathsAQARevision Notes2. Algebra & FunctionsPolynomials & Factor TheoremApplications of Factor Theorem

Applications of Factor Theorem (AQA GCSE Further Maths)

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Applications of Factor Theorem

What is factorising by inspection?

  • Factorising by inspection means expanding brackets in your head to find only one possibility
  • If (2x + 1) is a factor of 2x2 + 7x + 3, then...
    • 2 x squared plus 7 x plus 3 ≡ open parentheses 2 x plus 1 close parentheses open parentheses... plus... close parentheses
    • by inspection, the only possibility is open parentheses 2 x plus 1 close parentheses open parentheses x plus 3 close parentheses
      • this gives 2x2 and +3
  • The answer can be written down with no working

How do I factorise a cubic when one linear factor is given?

  • If you know a linear factor of a cubic expression, you can use that... 
    • ... "cubic expression" ≡ "linear factor" × "quadratic factor"
  • If (x - 2) is a factor of 2x3 + 7x2 - 17x - 10,  
    • 2x3 + 7x2 - 17x - 10 ≡ (x - 2)(ax2 + bx + c) 
    • By inspection, a = 2 and c = 5 (this give 2x3 and -10)
    • Find b by equating coefficients of x2 
      • 7x2 on the left
      • negative 2 cross times 2 x squared and x cross times b x on the right (search for x2 terms made up of something from the first bracket and something from the second bracket)
      • this gives negative 4 x squared plus b x squared on the right
      • b must be 11
    • 2x3 + 7x2 - 17x - 10 ≡ (x - 2)(2x2 + 11x + 5)
    • The final step is to factorise the quadratic factor
      • 2x2 + 11x - 5 ≡ (2x + 1)(x + 5)
      • so the cubic 2x3 + 7x2 - 17x - 10 factorises to (x - 2)(2x + 1)(x + 5)

How do I factorise a cubic without knowing any linear factors?

  • Find a linear factor using the Factor Theorem (then use the method above)
  • If (x - a) is a factor, then f(a) = 0, so test different values of a to find a factor
  • To find a linear factor of f(x) = x3 - 6x2 - x + 30, work out f(1), f(-1), f(2), f(-2), f(3), f(-3)... etc until you get f(a) = 0
      • straight f open parentheses 1 close parentheses equals 1 cubed minus 6 cross times 1 squared minus 1 plus 30 equals 24 straight f open parentheses negative 1 close parentheses equals open parentheses negative 1 close parentheses cubed minus 6 cross times open parentheses negative 1 close parentheses squared minus open parentheses negative 1 close parentheses plus 30 equals 24 straight f open parentheses 2 close parentheses equals 2 cubed minus 6 cross times 2 squared minus 2 plus 30 equals 12 straight f open parentheses negative 2 close parentheses equals open parentheses negative 2 close parentheses cubed minus 6 cross times open parentheses negative 2 close parentheses squared minus open parentheses negative 2 close parentheses plus 30 equals 0
      • the first zero came from f(-2) = 0 so, by the Factor Theorem, (x + 2) is a factor
    • You only need to test ± whole numbers that divide "30" (the number at the end of the cubic)
  • Cubics with a number in front of x3, such as f(x) = 3x3 + 4x2 - 5x - 2, are harder
    • try factors of (3x + 1), (3x - 1), (3x + 2), (3x - 2), ... as well as (x + 1), (x - 1), (x + 2), (x - 2)...

How do I solve a cubic equation?

  • Factorise the cubic using the method above, then set each bracket equal to zero and solve for x
  • To solve 3x3 + 4x2 - 5x - 2 = 0
    • f(1) = 0 so (x - 1) is a factor, factorising to (x - 1)(x + 2)(3x + 1) = 0
    • Solve each bracket equal to zero
      • x - 1 = 0 gives x = 1
      • x + 2 = 0 gives x = -2
      • 3x + 1 = 0 gives x equals negative 1 third
    • the solutions are x equals negative 2 comma space minus 1 third space and space 1

Exam Tip

  • Beware of trying to find all three linear factors by just testing numbers
    • you could find f(1) = 0, f(-1) = 0 and f(2) = 0 from straight f open parentheses x close parentheses equals 2 x cubed minus 4 x squared minus 2 x plus 4 and think it factorises to (x - 1)(x + 1)(x - 2) but it doesn't (expand and check)
    • some cubics only have one factor (so you'd be testing an infinite number of other integers trying to find non-existent factors!)

Worked example

Solve x cubed minus 5 x squared minus 16 x plus 80 equals 0

Set the polynomial equal to straight f open parentheses x close parentheses and find the first linear factor by testing positive and negative factors of 80, starting with the smallest values.

Test f(1) and f(-1).

table row cell straight f open parentheses 1 close parentheses space end cell equals cell space open parentheses 1 close parentheses cubed minus 5 open parentheses 1 close parentheses squared minus 16 open parentheses 1 close parentheses plus 80 space equals space 1 minus 5 minus 16 plus 80 space equals space 60 end cell row cell straight f open parentheses negative 1 close parentheses space end cell equals cell space open parentheses negative 1 close parentheses cubed minus 5 open parentheses negative 1 close parentheses squared minus 16 open parentheses negative 1 close parentheses plus 80 space equals space minus 1 minus 5 plus 16 plus 80 space equals space 90 end cell end table

Test f(2) and f(-2).

table row cell straight f open parentheses 2 close parentheses space end cell equals cell space open parentheses 2 close parentheses cubed minus 5 open parentheses 2 close parentheses squared minus 16 open parentheses 2 close parentheses plus 80 space equals space 8 minus 20 minus 32 plus 80 space equals space 36 end cell row cell straight f open parentheses negative 2 close parentheses space end cell equals cell space open parentheses negative 2 close parentheses cubed minus 5 open parentheses negative 2 close parentheses squared minus 16 open parentheses negative 2 close parentheses plus 80 space equals space minus 8 minus 20 plus 32 plus 80 space equals space 84 end cell end table

 

Test f(4) and f(-4), there is no need to test f(3) and f(-3) as 3 is not a factor of 80.

table row cell straight f open parentheses 4 close parentheses space end cell equals cell space open parentheses 4 close parentheses cubed minus 5 open parentheses 4 close parentheses squared minus 16 open parentheses 4 close parentheses plus 80 space equals space 64 minus 80 minus 64 plus 80 space equals space 0 end cell end table

The linear factor is found so there is no need to test anymore. 

straight f open parentheses 4 close parentheses space equals space 0 spaceso  x space minus space 4 is a factor.

table row cell straight f open parentheses x close parentheses space end cell equals cell space open parentheses x space minus space 4 close parentheses open parentheses a x squared space plus space b x space plus space c close parentheses space space end cell end table

Set the two expressions equal to each other and equate coefficients.

table row cell x cubed space minus space 5 x squared space minus 16 x space plus space 80 space end cell equals cell space open parentheses x space minus space 4 close parentheses open parentheses a x squared space plus space b x space plus space c close parentheses space space end cell end table

Equating the first terms, a x cubed space equals space x cubed therefore  a space equals space 1.

table row cell x cubed space minus space 5 x squared space minus 16 x space plus space 80 space end cell equals cell space open parentheses x space minus space 4 close parentheses open parentheses x squared space plus space b x space plus space c close parentheses space space end cell end table

Equating the last (constant) terms, negative 4 c space equals space 80 therefore  c space equals fraction numerator space 80 over denominator negative 4 end fraction space equals space minus 20.

table row cell straight f open parentheses x close parentheses space end cell equals cell space open parentheses x space minus space 4 close parentheses open parentheses x squared space plus space b x minus 20 close parentheses space space end cell end table

Consider the x squared terms by multiplying out these parts.

table row cell negative space 5 x squared space end cell equals cell space minus 4 a x squared plus space b x to the power of 2 space end exponent end cell end table

Substitute a space equals space 1 and solve for b.

table row cell negative space 5 space end cell equals cell space minus 4 space plus space b end cell row cell b space end cell equals cell space minus 1 end cell end table

Substitute in and factorise the quadratic.

table row cell straight f open parentheses x close parentheses space end cell equals cell space open parentheses x space minus space 4 close parentheses open parentheses x squared space minus space x minus 20 close parentheses space end cell row blank equals cell space open parentheses x space minus space 4 close parentheses open parentheses x space plus space 4 close parentheses open parentheses x space minus space 5 close parentheses space end cell end table

The solutions are the opposite signs of each factor.

bold italic x bold space bold equals bold space bold minus bold 4 bold comma bold space bold space bold space bold italic x bold space bold equals bold space bold 4 bold comma bold space bold space bold space bold italic x bold space bold equals bold space bold 5

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    Mark

    Author: Mark

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.