1. AS-Level Maths: Core 2 for Edexcel
C2.1 Algebra and functions
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2. Contents Dividing polynomials Dividing polynomials
The Remainder Theorem
The Factor Theorem
Examination-style questions
Contents
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3. Multiplying polynomials
Two polynomials are multiplied together and the resulting polynomial is x3 + x2 – 10x + 8.
One of the polynomials is (x + 4). What is the other?
x + 4 is a linear polynomial. To obtain a cubic polynomial as required we need to multiply it by a quadratic of the form
ax2 + bx + c.
We can write (x + 4)(ax2 + bx + c) ≡ x3 + x2 – 10x + 8.
This is an example of an identity as shown by the symbol ≡.
An identity is true for all values of x.
This problem is another way of asking what is (x3 + x2 – 10x + 8) divided by (x + 4). The method show on the next two slides can therefore be used for dividing polynomials as long as the possibility of a remainder is taken into account.
The symbol ≡ means ‘is identically equal to’.
Since the expression on the left hand side is equivalent to the expression on the right hand side, the coefficients of x3, x2, x and the constant must be the same on both sides.

4. Equating coefficients
We can use the method of equating coefficients to find the values of a, b and c in the identity
(x + 4)(ax2 + bx + c) ≡ x3 + x2 – 10x + 8
Multiplying out:
ax3 + bx2 + cx + 4ax2 + 4bx + 4c ≡ x3 + x2 – 10x + 8
ax3 + (b + 4a)x2 + (c + 4b)x + 4c ≡ x3 + x2 – 10x + 8
Equating the coefficients of x3 gives
a = 1
This method is also called comparing coefficients.
Equating the coefficients of x2 gives
b + 4a = 1
But a = 1 so
b + 4 = 1
b = –3

5. Equating coefficients
Equating the coefficients of x gives
c + 4b = –10
But b = –3 so
c – 12 = –10
c = 2
We can equate the constants to check this value:
4c = 8
c = 2
Substituting a = 1, b = –3 and c = 2 into
Conclude that x3 + x2 – 10x + 8 divided by x + 4 is x2 – 3x + 2.
(x + 4)(ax2 + bx + c) ≡ x3 + x2 – 10x + 8
Gives the solution
(x + 4)(x2 – 3x + 2) = x3 + x2 – 10x + 8
What is x3 + x2 – 10x + 8 divided by x + 4?

6. Dividing polynomials
Suppose we want to divide one polynomial f(x) by another polynomial of lower order g(x).
There are two possibilities. Either:
g(x) will leave a remainder when divided into f(x).
g(x) will divide exactly into f(x). In this case, g(x) is a factor of f(x) and the remainder is 0.
We can use either of two methods to divide one polynomial by another. These are by:
With practice, simple polynomials can also be divided by inspection.
using long division, or
writing an identity and equating coefficients.

7. Dividing polynomials by long division
Using long division
The method of long division used for numbers can be applied to the division of polynomial functions.
Let’s start by looking at the method for numbers.
For example, we can divide 5482 by 15 as follows: 5 15 4 8 2
This tells us that 15 divides into times, leaving a remainder of 7. 3 6 5 – 4 5
We can write
It is possible that some students have not come across the method of long division in the past. They may have used an alternative method such as subtracting multiples or ‘chunking’. If this is the case the method can be avoided by using the method of equating coefficients, as shown later.
If required, we could extend the method of long division to give the answer to a given number of decimal places. However, when examining the method to apply it to polynomials, an integer remainder is more desirable.
Point out that when two numbers or, indeed, two polynomial expressions are divided, we can write the result in the form: dividend = divisor × quotient + remainder. In the case where the divisor is a factor of the dividend, the remainder is 0 and so we simply have: dividend = divisor × quotient. 9 8
5482 ÷ 15 = 365 remainder 7 – 9 8 2
or = 15 × – 7 5 7
The
dividend
The
divisor =
The
quotient ×
The
remainder +

8. Dividing polynomials by long division
We can use the same method to divide polynomials.
For example:
What is f(x) = x3 – x2 – 7x + 3 divided by x – 3?
This tells us that x3 – x2 – 7x + 3 divided by x – 3 is x2 + 2x – 1.
x3 – x2 – 7x + 3
x – 3 x2
+ 2x
– 1
The remainder is 0 and so x – 3
is a factor of f(x).
x3 – 3x2
2x2
– 7x
We can write
2x2 – 6x
Talk through each step in the calculation explaining that we are choosing what value to multiply the divisor by to eliminate each term in x starting with the one of highest degree.
– x
+ 3
– x + 3 or
x3 – x2 – 7x + 3 =(x – 3)(x2 + 2x – 1)

9. Dividing polynomials by long division
Here is another example:
What is f(x) = 2x3 – 3x2 + 1 divided by x – 2?
This tells us that 2x3 – 3x2 + 1 divided by x – 2 is 2x2 + x + 2 remainder 5.
2x3 – 3x2 + 0x + 1
x – 2
2x2
+ x
+ 2
2x3 – 4x2
There is a remainder and so x – 2
is not a factor of f(x). x2
+ 0x
We can write
x2 – 2x
Talk through the example, drawing particular attention to the fact that a term in 0x has been added to the expression so that each column contains a different power of x. 2x
+ 1
2x – 4 or 5
2x3 – 3x2 + 1 =(x – 2)(2x2 + x + 2) + 5

10. Dividing polynomials Using the method of equating coefficients
Polynomials can also be divided by constructing an appropriate identity and equating the coefficients. For example:
What is f(x) = 3x2 + 11x – 8 divided by x + 5?
We can write f(x) = 3x2 + 11x – 8 in terms of a quotient and a remainder as follows:
3x2 + 11x – 8 = (x + 5)(quotient) + (remainder)
To obtain a quadratic polynomial the quotient must be a linear polynomial of the form ax + b.
This method has the advantage of avoiding long division. With practice, it can be done in fewer steps.
Point out that if x + 5 is a factor of 3x2 + 11x – 8 we would expect the remainder to be 0.
We can write the following identity:
3x2 + 11x – 8 ≡ (x + 5)(ax + b) + r

11. 3x2 + 11x – 8 divided by x + 5 is 3x – 4 remainder 12.
Dividing polynomials
Expanding:
3x2 + 11x – 8 ≡ ax2 + bx + 5ax + 5b + r
≡ ax2 + (b + 5a)x + 5b + r
Equating the coefficients of x2:
a = 3
Equating the coefficients of x:
b + 5a = 11
But a = 3 so
b + 15 = 11
b = –4
Equating the constants:
5b + r = –8
But b = –4 so
–20 + r = –8
r = 12
We can use these values to write
3x2 + 11x – 8 ≡ (x + 5)(3x – 4) + 12 So
3x2 + 11x – 8 divided by x + 5 is 3x – 4 remainder 12.

12. Contents The Remainder Theorem Dividing polynomials
The Factor Theorem
Examination-style questions
Contents
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13. The Remainder Theorem
Consider again the example where f(x) = 3x2 + 11x – 8 divided by x + 5.
Find the value of f(–5).
f(–5) = 3(–5)2 + 11(–5) – 8
= 75 – 55 – 8
= 12
Can you explain why this number should be the same as the remainder when f(x) = 3x2 + 11x – 8 divided by x + 5?
Discuss the fact that when a polynomial f(x) is divided by an expression of the form (x – a) we can find the remainder by putting x equal to a.
This is the remainder theorem and is given in its general form on the next slide.
f(x) = (x + 5)(quotient) + (remainder)
If x = –5 then f(–5) = (–5 + 5)(quotient) + (remainder)
= 0 + (remainder)
So when f(x) is divided by (x + 5), f(–5) = the remainder.

14. The Remainder Theorem
When a polynomial f(x) is divided by (x – a), the remainder is f(a).
This is called the Remainder Theorem.
We can use this theorem to find the remainder when a polynomial is divided by an expression of the form (x – a).
For example:
Find the remainder when the polynomial
f(x) = x3 – 3x2 – 8x + 5 is divided by (x + 2).
Warn students to be careful with signs when substituting negative numbers. In particular, they should remember that a negative number raised to an odd power always gives an negative number, and that a negative number raised to an even power always gives a positive number.
Using the Remainder Theorem, the remainder is given by f(–2).
f(–2) = (–2)3 – 3(–2)2 – 8(–2) + 5
= –8 –
= 1

15. Contents The Factor Theorem Dividing polynomials The Remainder Theorem
Examination-style questions
Contents
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16. The Factor Theorem
Suppose that when a polynomial f(x) is divided by an expression of the form (x – a) the remainder is 0.
What can you conclude about (x – a)?
If the remainder f(a) is 0 then (x – a) is a factor of f(x).
This is the Factor Theorem:
If (x – a) is a factor of a polynomial f(x) then f(a) = 0.
The factor theorem is a special case of the remainder theorem: if f(a) = 0, then the remainder is zero and so (x – a) is a factor.
Explain that the first form of the factor theorem can be used to check whether a given expression of the form (x – a) is a factor of a given polynomial.
The converse form of the factor theorem can be used to find factors of a given polynomial by choosing values of x until we find a value that makes the polynomial equal to 0.
The converse is also true:
If f(a) = 0 then (x – a) is a factor of a polynomial f(x).

17. The Factor Theorem Use the Factor Theorem to show that
(x + 2) is a factor of f(x) = 3x2 + 5x – 2.
Hence or otherwise, factorize f(x).
(x + 2) is a factor of 3x2 + 5x – 2 if f(–2) = 0
f(–2) = 3(–2)2 + 5(–2) – 2
= 12 – 10 – 2
= 0 as required.
In this example, f(x) can be factorized by inspection because it is quadratic. (Students should have had lots of practice factorizing quadratics).
If f(x) had been of a higher degree than 0, as shown in the next example, we could have factorized it by the method of equating coefficients by using long division.
We can write
3x2 + 5x – 2 = (x + 2)(ax + b)
By inspection
a = 3 and b = –1
And so
3x2 + 5x – 2 = (x + 2)(3x – 1)

18. Factorizing polynomials
The Factor Theorem can be used to factorize polynomials by systematically looking for values of x that will make the polynomial equal to 0. For example:
Factorize the cubic polynomial x3 – 3x2 – 6x + 8.
Let f(x) = x3 – 3x2 – 6x + 8.
f(x) has a constant term of 8. So possible factors of f(x) are:
(x ± 1),
(x ± 2),
(x ± 4)
(x ± 8) or
f(1) = 1 – 3 – = 0
 (x – 1) is a factor of f(x).
f(–1) = –1 – ≠ 0
 (x + 1) is not a factor of f(x).
f(2) = 8 – 12 – ≠ 0
 (x – 2) is not a factor of f(x).

19. Factorizing polynomials
 (x + 2) is a factor of f(x).
f(4) = 64 – 48 – = 0
 (x – 4) is a factor of f(x).
We have found three factors and so we can stop.
The given polynomial can therefore be fully factorized as:
x3 – 3x2 – 6x + 8 = (x – 1)(x + 2)(x – 4)
Factorize f(x) = x3 + 1
f(x) has a constant term of 1 so the only possible factors of f(x) are (x – 1) or (x + 1).
f(1) = ≠ 0
 (x – 1) is not a factor of f(x).
f(–1) = (–1)3 + 1 = 0
 (x + 1) is a factor of f(x).

20. The factor theorem
We don’t know any other factors but we do know that the expression x + 1 must be multiplied by a quadratic expression to give x We can therefore write
x3 + 1 = (x + 1)(ax2 + bx + c)
We can see immediately that a = 1 and c = 1 so
x3 + 1 = (x + 1)(x2 + bx + 1)
= x3 + bx2 + x + x2 + bx + 1
= x3 + (b + 1)x2 + (b + 1)x + 1
Equating coefficients of x2 gives
The quotient could also have been found by dividing x3 + 1 by x + 1 using long division. However, for this example, the method of equating coefficients is much quicker.
Establish, by finding the discriminant of x2 – x + 1, that x3 + 1 cannot be factorized any further.
b + 1 = 0
b = –1
So x3 + 1 can be fully factorized as
x3 + 1 = (x + 1)(x2 – x + 1)

21. Extending the remainder and factor theorems
Suppose we want to know the remainder when a polynomial f(x) is divided by an expression of the form (ax – b). We can write
f(x) = (ax – b)(quotient) + (remainder)
We can eliminate the quotient by choosing x so that
ax – b = 0
ax = b
So, in general:
When a polynomial f(x) is divided by (ax – b),
the remainder is

22. Extending the remainder and factor theorems
Find the remainder when the polynomial
f(x) = 2x3 – 6x + 1 is divided by (2x – 1).
Using the Remainder Theorem, the remainder is given by
We can similarly extend the Factor Theorem to include factors of the form (ax – b).
Explain the use of the  symbol to mean ‘implies and is implied by’ or ‘if and only if’.
We can write:
Where  means ‘implies and is implied by’.

23. Examination-style questions
Dividing polynomials
The Remainder Theorem
The Factor Theorem
Examination-style questions
Contents
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24. Examination-style question
p(x) = 4x3 + ax2 + bx – 15
p(x) leaves a remainder of 27 when divided by x – 2.
x + 1 is a factor of p(x).
Find the values of a and b and hence write p(x) in full.
Express p(x) as a product of linear factors.
Sketch the graph of y = p(x) clearly indicating where the graph crosses the coordinate axes.
a) Using the remainder theorem:
p(2) = a + 2b – 15 = 27
4a + 2b = 10
2a + b = 50 1

25. Examination-style question
Using the factor theorem:
p(–1) = –4 + a – b – 15 = 0
a – b = 19 1
Adding equations and : 1 2
3a = 24
a = 8
b = –11
So p(x) can be written in full as
p(x) = 4x3 + 8x2 – 11x – 15

26. Examination-style question
b) We know that x + 1 is a factor of p(x) so we can write
4x3 + 8x2 – 11x – 15 ≡ (x + 1)(ax2 + bx + c)
By inspection
a = 4 and c = –15
Equating the coefficients of x:
–11 = c + b
–11 = –15 + b
b = 4
p(x) = 4x3 + 8x2 – 11x – 15 = (x + 1)(4x2 + 4x – 15) So
Factorizing:
p(x) = 4x3 + 8x2 – 11x – 15 = (x + 1)(2x + 5)(2x – 3)

27. Examination-style question
c) The graph of y = p(x) crosses the x-axis when
4x3 + 8x2 – 11x – 15 = 0
(x + 1)(2x + 5)(2x – 3) = 0
That is when x = –1, x = –2.5, and x = 1.5
The graph of y = p(x) crosses the y-axis when
y = – 15 x y
–15
1.5 –1
–2.5
Also the coefficient of x3 is positive so the graph is -shaped: