Applications of Factor Theorem (AQA GCSE Further Maths)
Revision Note
Written by: Mark Curtis
Reviewed by: Dan Finlay
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...
≡
by inspection, the only possibility is
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
and on the right (search for x2 terms made up of something from the first bracket and something from the second bracket)
this gives 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
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
the solutions are
Examiner Tips and Tricks
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 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
Set the polynomial equal to 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).
Test f(2) and f(-2).
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.
The linear factor is found so there is no need to test anymore.
so is a factor.
Set the two expressions equal to each other and equate coefficients.
Equating the first terms, therefore .
Equating the last (constant) terms, therefore .
Consider the terms by multiplying out these parts.
Substitute and solve for .
Substitute in and factorise the quadratic.
The solutions are the opposite signs of each factor.
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