1. The document discusses dividing polynomials by using long division or writing an identity and equating coefficients. 2. It provides examples of using both methods to divide polynomials and determine quotients and remainders. 3. The Remainder Theorem is introduced, which states that when a polynomial f(x) is divided by (x - a), the remainder is equal to the value of f(a).

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1. Algebra and
Function
Part 1

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SINGING
INDONESIA
RAYA
01
INTRODUCTION
OF THE TOPIC
03
OBJECTIVE
02
EXAMPLES
04
05 Discussion

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4. 4 of 18
Objective:

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Contents
© Boardworks Ltd 2005
5 of 27
Dividing polynomials
The Remainder Theorem
The Factor Theorem
Examination-style questions
Dividing polynomials

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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.
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.

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

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Equating coefficients
Equating the coefficients of x gives
We can equate the constants to check this value:
4c = 8
Substituting a = 1, b = –3 and c = 2 into
c + 4b = –10
But b = –3 so c – 12 = –10
c = 2
c = 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?

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Dividing polynomials
Suppose we want to divide one polynomial f(x) by another
polynomial of lower order g(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.
There are two possibilities. Either:
g(x) will leave a remainder when divided into f(x).
We can use either of two methods to divide one polynomial by
another. These are by:
using long division, or
writing an identity and equating coefficients.

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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:
This tells us that 15 divides into 5482
365 times, leaving a remainder of 7.
We can write
or 5482 = 15 × 365 + 7
The
dividend
5482 ÷ 15 = 365 remainder 7
The
divisor
=
The
quotient
×
The
remainder
+
5
15 4 8 2
3
– 4 5
9 8
6
9 0
–
8 2
5
7 5
–
7

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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?
x3 – x2 – 7x + 3
x – 3
x3 – 3x2
2x2 – 7x
+ 2x
2x2 – 6x
– x + 3
– 1
– x + 3
0
This tells us that x3 – x2 – 7x + 3
divided by x – 3 is x2 + 2x – 1.
The remainder is 0 and so x – 3
is a factor of f(x).
We can write
x3 – x2 – 7x + 3 =(x – 3)(x2 + 2x – 1)
3 2
2
7 +3
= + 2 1
3
x x x
x x
x
 


x2
or

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Dividing polynomials by long division
Here is another example:
What is f(x) = 2x3 – 3x2 + 1 divided by x – 2?
2x3 – 3x2 + 0x + 1
x – 2
2x3 – 4x2
x2 + 0x
+ x
x2 – 2x
2x + 1
+ 2
2x – 4
5
This tells us that 2x3 – 3x2 + 1
divided by x – 2 is 2x2 + x + 2
remainder 5.
There is a remainder and so x – 2
is not a factor of f(x).
We can write
2x3 – 3x2 + 1 =(x – 2)(2x2 + x + 2) + 5
2x2
3 2
2
2 3 1
2
2
5
2
2
x x
x
x x
x
 
 
 
 
or

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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.
We can write the following identity:
3x2 + 11x – 8 ≡ (x + 5)(ax + b) + r

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

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Contents
© Boardworks Ltd 2005
15 of 27
The Remainder Theorem
Dividing polynomials
The Remainder Theorem
The Factor Theorem
Examination-style questions

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The Remainder Theorem
Consider again the example where f(x) = 3x2 + 11x – 8 divided
by x + 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?
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.
Find the value of f(–5).

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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).
Using the Remainder Theorem, the remainder is given by f(–2).
f(–2) = (–2)3 – 3(–2)2 – 8(–2) + 5
= –8 – 12 + 16 + 5
= 1

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Thank you