Download AQA Core 1 Polynomials Section 2: The factor

AQA Core 1 Polynomials
Section 2: The factor and remainder theorems
Notes and Examples
These notes contain subsections on
 The factor theorem
 Dividing polynomials
 The remainder theorem
The factor theorem
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
This idea can be extended to other polynomials such as cubics.
For example, for the cubic function g( x)  ( x 1)( x  2)( x  3) , g(1) = 0, g(-2) = 0
and g(3) = 0.
(x – 1) is a factor of g(x)  g(1) = 0
(x + 2) is a factor of g(x)  g(-2) = 0
(x – 3) is a factor of g(x)  g(3) = 0
In general, the factor theorem states that:
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).
You can use the factor theorem to solve cubic and higher order equations.
These are the steps you need to take to solve a cubic equation of the form
f(x) = 0, where f(x) is a cubic function:
 First, work out f(x) for different values of x until you find one for which
f(x) = 0.
 Using the factor theorem, you now know one factor of the function.
 Divide the function by this linear factor. You can now express the
function as the product of the linear factor and a quadratic factor.
 Factorise the quadratic factor, if possible.
 If the quadratic factor does factorise, you now have three linear factors,
from which you can deduce the three solutions to the equation.
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AQA C1 Polynomials 2 Notes and Examples

If the quadratic factor does not factorise, you can use the quadratic
formula to find the two further solutions, if they exist.
For higher order equations, you will have to find more than one factor by trial
and error, and you will have to divide more than once.
The following example shows how this method works.
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).
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.
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AQA C1 Polynomials 2 Notes and Examples
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.
You can see some different approaches using the Flash resources
Polynomial division by inspection, Polynomial division – box method
(no remainder) and Factorising a cubic, and the PowerPoint presentation
Factorising polynomials. The Geogebra resource Polynomial division
shows the box method.
You can also look at the Solving cubics video.
The Flash resource Sketching factorised cubics and the Mathcentre video
Polynomial functions may also be useful.
For some additional practice, try the interactive questions Sketching
polynomial curves and Finding a polynomial from its roots.
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
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AQA C1 Polynomials 2 Notes and Examples
= 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.
Dividing polynomials
When you divide one polynomial by another, it may divide exactly, or there
may be a remainder, just as in arithmetic.
For example:
26 is called the dividend,
6 is called the divisor,
4 is the quotient
and 2 is the remainder.
26  6 = 4 remainder 2
You could rewrite this statement as:
26 = 6  4 + 2
This rearrangement helps to give one method of dividing polynomials.
When working with polynomials, remember the following points:
 If you are dividing by a linear expression, the quotient is of order one
less than the dividend (e.g. for a quartic, the quotient is cubic) and the
remainder, if any, is a constant term.
 If you are dividing by a quadratic term, the quotient is of order two less
than the dividend (e.g. for a quartic, the quotient is quadratic) and the
remainder, if any, could be linear or a constant term.
 This idea can be extended to a polynomial divisor of any order.
Most, if not all, of the examples you meet will only involve dividing by a linear
expression.
When you divide a cubic expression
by a linear expression, as in this
example, the quotient is a quadratic
expression and the remainder, if
any, is a constant term.
Example 3
Divide 2x³ + 3x² - x + 1 by x + 2
Solution
Let the quotient be ax² + bx + c and the remainder be d.
2x³ + 3x² – x + 1
= (x + 2)(ax² + bx + c) + d
= x(ax² + bx + c) + 2(ax² + bx + c) + d
= ax³ + bx² + cx + 2ax² + 2bx + 2c + d
= ax³ + (b + 2a)x² + (c + 2b)x + 2c + d
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AQA C1 Polynomials 2 Notes and Examples
Equating coefficients of x³
Equating coefficients of x²
Equating coefficients of x
Equating constant terms
a=2
 b + 2a = 3  b + 4 = 3  b = -1
 c + 2b = -1  c – 2 = -1  c = 1
 2c + d = 1  2 + d = 1  d = -1
2x³ + 3x² – x + 1 = (x + 2)(2x² – x + 1) – 1
The quotient is 2x² – x + 1 and the remainder is -1.
Don’t be put off by a negative remainder, as in this example – this is quite
acceptable in polynomial division!
There are other approaches to polynomial division. The PowerPoint
presentation Dividing polynomials shows (1) algebraic long division and (2)
polynomial division by inspection (i.e. in your head). You can also look at
more examples using the Flash resources Polynomial division (long
division) and Polynomial division – box method (remainder).
There is also a Polynomial division video – this deals with the long division
method only.
For some extra practice in examples like the one above, try the interactive
resource Dividing polynomials.
The remainder theorem
The factor theorem is a special case of the remainder theorem.
When a polynomial f(x) is divided by a linear expression (x – a), then if (x – a)
is not a factor of f(x), then there will be a remainder.
f(x) = (x – a)  quotient + remainder
Substituting x = a into this expression gives:
i.e.
f(a) = 0  quotient + remainder
f(a) = remainder
The remainder theorem says:
For a polynomial f(x), f(a) is the remainder when f(x) is divided by  x  a  .
f ( x)remainder
  x  a  g( is
x) 
f (a) then
.
If the
zero,
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.
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AQA C1 Polynomials 2 Notes and Examples
Example 4
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 5
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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