MEI Core 1 Polynomials
Section 2: The factor and remainder theorems
Notes and Examples
These notes contain subsections on
 The factor theorem
 The remainder theorem
The factor theorem
If  x  a  is a factor of f(x), then f(a) = 0 and x = a is a root of the equation f(x) = 0.
Conversely, if f(a) = 0, then  x  a  is a factor of f(x).
The factor theorem probably won’t come as any great surprise to you. After
all, you already know that you can solve some quadratics by factorising them.
e.g.
to solve the quadratic equation
you factorise:
and deduce the solutions
x² + 3x – 10 = 0
(x + 5)(x – 2) = 0
x = -5 and x = 2
Clearly, for f(x) = x² + 3x – 10, f(-5) = 0 and f(2) = 0.
(x + 5) is a factor of f(x)  f(-5) = 0
(x – 2) is a factor of f(x)  f(2) = 0
The factor theorem simply extends this idea to other polynomials such as
cubics, and provides a method for solving cubics and higher polynomials.
Example 1
(i) Solve the equation x³ + 2x² – 5x – 6 = 0
(ii) Sketch the graph of y = x³ + 2x² – 5x – 6
Solution
(i)
The first step is to find one solution by trial and error.
If there is an integer solution x = a, then by the factor theorem
(x – a) must be a factor of x³ + 2x² – 5x – 6. So a must be a
factor of 6. a could therefore be 1, -1, 2, -2, 3, -3, 6 or –6.
Let f(x) = x³ + 2x² – 5x – 6
You need to find a value of
x for which f(x) = 0.
f(1) = 1 + 2 – 5 – 6 = -8
f(-1) = -1 + 2 + 5 – 6 = 0
f(-1) = 0 so by the factor theorem x + 1 is a factor of f(x).
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The next step is to factorise f(x) into the linear factor
x + 1 and a quadratic factor.
x³ + 2x² – 5x – 6 = (x + 1)  quadratic factor.
Let the quadratic factor be ax² + bx + c.
x³ + 2x² – 5x – 6 = (x + 1)(ax² + bx + c)
= ax³ + bx² + cx + ax² + bx + c
= ax³ + (a + b)x² + (b + c)x + c
Multiply out the
brackets
Equating coefficients of x³  a = 1
Equating constant term  c = -6
Equating coefficients of x²  a + b = 2  b = 1
Check coefficient of x: b + c = 1 – 6 = -5
x³ + 2x² – 5x – 6 = (x + 1)(x² + x – 6)
= (x + 1)(x – 2)(x + 3)
Factorise the
quadratic factor
The solutions of the equation are x = -1, x = 2 and x = -3.
(ii) Part (i) shows that the graph of y = x³ + 2x² – 5x – 6 crosses the x-axis at (-3, 0),
(-1, 0) and (2, 0). By putting x = 0 you can see that it crosses the y-axis at (0, -6).
This information allows you to sketch the graph.
In the example above, the quadratic factor was found by equating
coefficients. There are a number of other methods of finding this quadratic
factor. You can do it by polynomial division, or by inspection (which means
doing it in your head). Which one you use is really down to personal
preference.
The PowerPoint presentation Factorising polynomials shows these
alternative methods.
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MEI C1 Polynomials 2 Notes and Examples
You can look at some more examples of factorising cubics using the Flash
resource Factorising a cubic.
You can also look at the Solving cubics video.
Example 2
f(x) = 2x³ + px² + 5x – 6 has a factor x – 2.
Find the value of p and hence factorise f(x) as far as possible.
Solution
x – 2 is a factor of f(x)  f(2) = 0
f(2) = 16 + 4p + 10 – 6 = 20 + 4p
20 + 4p = 0  p = -5
f(x) = 2x³ - 5x² + 5x – 6
2x³ - 5x² + 5x – 6
= (x – 2)(ax² + bx + c)
= ax³ + bx² + cx – 2ax² - 2bx – 2c
= ax³ + (b – 2a)x² + (c – 2b)x – 2c
Equating coefficients of x³  a = 2
Equating constant terms  -2c = -6  c = 3
Equating coefficients of x²  b – 2a = -5  b – 4 = -5  b = -1
Check coefficient of x: c – 2b = 3 + 2 = 5
2x³ - 5x² + 5x – 6 = (x – 2)(2x² – x + 3)
The discriminant of the quadratic factor is (-1)² - 4 2  3 = 1 – 24 = -23
As this is negative, the quadratic factor cannot be factorised further.
The remainder theorem
The factor theorem is a special case of the remainder theorem.
The remainder theorem says:
For a polynomial f(x), f(a) is the remainder when f(x) is divided by  x  a  .
f ( x)   x  a  g( x)  f (a)
If the remainder is zero, then x – a is a factor of f(x).
So if x – a is a factor of f(x), then f(a) = 0.
This is the factor theorem.
Don’t forget about the remainder theorem! When a remainder is asked for in
examination questions, there are usually quite a number of candidates who
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divide out to find it. This takes longer and you are much more likely to make
an error. The remainder theorem is quick and straightforward to use.
Example 3
Find the remainder when f(x) = x4 – 3x³ + x² – 4 is divided by
(i) x – 2
(ii) x + 1
Solution
(i) By the remainder theorem the remainder is f(2)
f(2) = 24 – 32³ + 2² - 4 = 16 – 24 + 4 – 4 = -8
Remainder = -8
(ii) By the remainder theorem the remainder is f(-1)
f(-1) = (-1)4 – 3(-1)³ + (-1)² – 4 = 1 + 3 + 1 – 4 = 1
Remainder = 1
You can look at similar examples using the Flash resource The remainder
theorem.
For extra practice in examples like the one above, try the interactive
questions Finding a remainder.
Example 4
f(x) = 2x³ + ax² + bx + 1
When f(x) is divided by x –1 the remainder is 7
When f(x) is divided by x + 3 the remainder is -5
Find the values of a and b.
Solution
f(1) = 7
f(-3) = -5
Adding:
2+a+b+1=7
a+b=4
 2(-3)³ + (-3)²a – 3b + 1 = -5
 -54 + 9a – 3b + 1 = -5
 9a – 3b = 48
 3a – b = 16
a +b=4
3a – b = 16
4a
= 20
a = 5, b = -1
For extra practice in examples like the one above, try the interactive
questions Finding a polynomial, given the remainder.
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