Transcript C2. 1 Algebra and functions
AS-Level Maths: Core 2
for Edexcel
C2.1 Algebra and functions
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Dividing polynomials
Dividing polynomials The Remainder Theorem The Factor Theorem Examination-style questions
Multiplying polynomials
Two polynomials are multiplied together and the resulting polynomial is
x
3 +
x
2 – 10
x
+ 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
ax
2 +
bx
+
c
.
We can write (
x
+ 4)(
ax
2 +
bx
+
c
) ≡
x
3 +
x
2 – 10
x
+ 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
x
3 ,
x
2 ,
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 (
x
+ 4)(
ax
2 +
bx
+
c
) ≡
x
3 +
x
2 – 10
x
+ 8 Multiplying out:
ax
3 +
bx
2 +
cx
+ 4
ax
2 + 4
bx
+ 4
c
≡
x
3 +
x
2 – 10
x
+ 8
ax
3 + (
b
+ 4
a
)
x
2 + (
c
+ 4
b
)
x
+ 4
c
≡
x
3 +
x
2 – 10
x
+ 8 Equating the coefficients of
x
3 gives
a
= 1 Equating the coefficients of
x
2 gives But
a
= 1 so
b
+ 4
a
= 1
b
+ 4 = 1
b
= –3 4 of 27 © Boardworks Ltd 2005
Equating coefficients
Equating the coefficients of
x
gives
c
+ 4
b
= –10 But
b
= –3 so
c
– 12 = –10
c
= 2 We can equate the constants to check this value: 4
c
= 8
c
= 2 Substituting
a
= 1,
b
= –3 and
c
= 2 into (
x
+ 4)(
ax
2 +
bx
+
c
) ≡
x
3 +
x
2 – 10
x
+ 8 Gives the solution (
x
+ 4) (
x
2 – 3
x
+ 2) =
x
3 +
x
2 – 10
x
+ 8 5 of 27 What is
x
3 +
x
2 – 10
x
+ 8 divided by
x
+ 4?
© Boardworks Ltd 2005
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: 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: 15 7 of 27 3 6 5 4 8 2 – 4 5 9 8 5 – 9 0 8 2 – 7 5 7 This tells us that 15 divides into 5482 365 times, leaving a remainder of 7 .
We can write 5482 ÷ 15 = 365 remainder 7
or
5482 = 15 × 365 + 7
The dividend = The divisor × The quotient + The remainder
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Dividing polynomials by long division
We can use the same method to divide polynomials. For example: What is
f
(
x
) =
x
3 –
x
2 – 7
x
+ 3 divided by
x
– 3?
x
– 3
x
2 + 2
x
– 1
x
3 –
x
3
– x
2 3
x
2 – 7
x
2
x
2 – 7
x +
3 2
x
2 – 6
x
–
x
+ 3 –
x
+ 3 0 This tells us that
x
3 divided by
x
– – 3 is
x
2
x
2 + 2 – 7
x x
+ 3 – 1 .
The remainder is 0 and so
x
– 3 is a factor of
f
(
x
).
We can write
x
3
x
2
x
3
x
3 –
x
2 =
x
2 + 2
x
1 or – 7
x
+ 3 =(
x
– 3)(
x
2 + 2
x
– 1 ) 8 of 27 © Boardworks Ltd 2005
Dividing polynomials by long division
Here is another example: What is
f
(
x
) = 2
x
3 – 3
x
2 + 1 divided by
x
– 2?
x
– 2 2
x
2 2
x
3 – 3
x
2 2
x
3
–
4
x
2 + + 0
x x
2 + 0
x x
2 – 2
x x
+ 2 + 1 2
x
+ 1 2
x
– 4 5 This tells us that 2
x
3 divided by
x
– 3
x
2 – 2 is 2
x
2 +
x
+ 1 + 2 remainder 5 .
There is a remainder and so
x
– 2 is not a factor of
f
(
x
).
We can write 2 2
x
3 3
x
2
x
2 1 2
x
2 or
x
3 2
x
5 2 – 3
x
2 + 1 =(
x
– 2)( 2
x
2 +
x
+ 2 ) + 5 9 of 27 © Boardworks Ltd 2005
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
) = 3
x
2 + 11
x
– 8 divided by
x
+ 5?
We can write
f
(
x
) = 3
x
2 + 11
x
remainder as follows: – 8 in terms of a quotient and a 3
x
2 + 11
x
– 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: 3
x
2 + 11
x
– 8 ≡ (
x
+ 5)(
ax
+
b
) +
r
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Dividing polynomials
Expanding: 3
x
2 + 11
x
– 8 ≡
ax
2 ≡
ax
2 +
bx
+ (
b
+ 5
ax
+ 5
b
+
r
+ 5
a
)
x
+ 5
b
+
r
Equating the coefficients of
x
2 :
a
= 3 Equating the coefficients of But
a
But
b
= 3 so Equating the constants: = –4 so
x
:
b
+ 5
a
= 11
b
+ 15 = 11
b
= –4 5
b
+
r
–20 +
r
= –8 = –8
r
= 12 We can use these values to write 3
x
2 + 11
x
– 8 ≡ (
x
+ 5)(3
x
– 4) + 12 So 3
x
2 + 11
x
– 8 divided by
x
+ 5 is 3
x
– 4 remainder 12 .
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The Remainder Theorem
Dividing polynomials The Remainder Theorem The Factor Theorem Examination-style questions
The Remainder Theorem
Consider again the example where
f
(
x
) = 3
x
2 + 11
x
– 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
) = 3
x
2 + 11
x
– 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. 13 of 27 © Boardworks Ltd 2005
The Remainder Theorem
When a polynomial
f
(
x
) is divided by (
x
– the remainder is
f
(
a
).
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:
f
( Find the remainder when the polynomial
x
) =
x
3 – 3
x
2 – 8
x
+ 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 14 of 27 © Boardworks Ltd 2005
The Factor Theorem
Dividing polynomials The Remainder Theorem The Factor Theorem Examination-style questions
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 converse is also true: If
f
(
a
) = 0 then (
x
–
a
) is a factor of a polynomial
f
(
x
).
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The Factor Theorem
( Use the Factor Theorem to show that
x
+ 2) is a factor of
f
(
x
) = 3
x
2 + 5
x
– 2.
Hence or otherwise, factorize
f
(
x
).
(
x
+ 2) is a factor of 3
x
2 + 5
x
– 2 if
f
( –2) = 0
f
( –2) = 3(–2) 2 + 5( –2) – 2 = 12 – 10 – 2 We can write By inspection And so = 0 3
x
2 + 5
x
– 2 = (
x
+ 2)(
ax
+
b
) as required.
a
= 3 and
b
= –1 3
x
2 + 5
x
– 2 = (
x
+ 2)(3
x
– 1) 17 of 27 © Boardworks Ltd 2005
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
x
3 – 3
x
2 – 6
x
+ 8.
Let
f
(
x
) =
x
3 – 3
x
2 – 6
x
+ 8.
f f f
(
x
) has a constant term of 8. So possible factors of
f
(
x
) are:
f
(1) = 1 – 3 – 6 + 8 = 0 ( –1) = –1 – 3 + 6 + 8 ≠ 0 (2) = 8 (
x
± 1), (
x
± 2), – 12 – 12 + 8 ≠ 0 (
x
± 4) or (
x
± 8) (
x
– 1) is a factor of
f
(
x
).
(
x
+ 1) is not a factor of
f
(
x
).
(
x
– 2) is not a factor of
f
(
x
).
18 of 27 © Boardworks Ltd 2005
Factorizing polynomials
f f
( –2) = – 8 – 12 + 12 + 8 = 0 (4) = 64 – 48 – 24 + 8 = 0 (
x
+ 2) is a factor of
f
(
x
).
(
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:
x
3 – 3
x
2 – 6
x
+ 8 = (
x
– 1)(
x
+ 2)(
x
– 4) Factorize
f
(
x
) =
x
3 + 1.
f
(
x
) has a constant term of 1 so the only possible factors of
f
(
x
) are (
x
– 1) or (
x
+ 1).
f f
( (1) = 1 + 1 –1) = (–1) 3 ≠ 0 + 1 = 0 (
x
– 1) is not a factor of
f
(
x
).
(
x
+ 1) is a factor of
f
(
x
).
19 of 27 © Boardworks Ltd 2005
The factor theorem
We don’t know any other factors but we do know that the expression
x
to give
x
3 + 1 must be multiplied by a quadratic expression + 1. We can therefore write
x
3 + 1 = (
x
+ 1)(
ax
2 +
bx
+
c
) We can see immediately that
a
= 1 and
c
= 1 so
x
3 + 1 = (
x
+ 1)(
x
2 +
bx
+ 1) =
x
3 +
bx
2 +
x
+
x
2 +
bx
+ 1 =
x
3 + (
b
+ 1)
x
2 + (
b
+ 1)
x
+ 1 Equating coefficients of
x
2 gives
b
+ 1 = 0
b
= –1 So
x
3 + 1 can be fully factorized as
x
3 + 1 = (
x
+ 1)(
x
2 –
x
+ 1) 20 of 27 © Boardworks Ltd 2005
Extending the remainder and factor theorems
f
Suppose we want to know the remainder when a polynomial (
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 So, in general:
ax
=
b b x
=
a
When a polynomial
f
(
x
) is divided by (
ax
the remainder is
f b
.
–
b
), 21 of 27 © Boardworks Ltd 2005
Extending the remainder and factor theorems
Find the remainder when the polynomial
f
(
x
) = 2
x
3 – 6
x
+ 1 is divided by (2
x
– 1).
Using the Remainder Theorem, the remainder is given by
f
( ) = 2( ) 3 2 2 = 1 4 3 +1 = 1 3 4 2
f
( ).
2 We can similarly extend the Factor Theorem to include factors of the form (
ax
–
b
).
We can write: 22 of 27
f b
= 0 (
ax
b
Where means ‘
implies and is implied by
’.
© Boardworks Ltd 2005
Examination-style questions
Dividing polynomials The Remainder Theorem The Factor Theorem Examination-style questions
Examination-style question
p
(
x
) = 4
x
3 +
ax
2 +
bx
– 15
p
(
x
) leaves a remainder of 27 when divided by
x
– 2.
x
+ 1 is a factor of
p
(
x
). a) Find the values of
a
and
b
and hence write
p
(
x
) in full.
b) Express
p
(
x
) as a product of linear factors.
c) Sketch the graph of
y
=
p
(
x
) clearly indicating where the graph crosses the coordinate axes. a) Using the remainder theorem:
p
(2) = 32 + 4
a
+ 2
b
– 15 = 27 4
a
+ 2
b
= 10 2
a
+
b
= 5
1
24 of 27 © Boardworks Ltd 2005
Examination-style question
Using the factor theorem:
p
( –1) = –4 +
a
–
b
– 15 = 0
a
–
b
= 19
1
3
a
= 24
a
= 8
b
= –11 So
p
(
x
) can be written in full as
p
(
x
) = 4
x
3 + 8
x
2 – 11
x
– 15 25 of 27 © Boardworks Ltd 2005
Examination-style question
b) We know that
x
+ 1 is a factor of
p
(
x
) so we can write 4
x
3 + 8
x
2 – 11
x
– 15 ≡ (
x
+ 1)(
ax
2 +
bx
+
c
) By inspection
a
= 4 and
c
= –15 Equating the coefficients of
x
: –11 =
c
+
b
–11 = –15 +
b
So
p
(
x
) = 4
x
3 + 8
x
2
b
= 4 – 11
x
– 15 = (
x
+ 1)(4
x
2 + 4
x
– 15) Factorizing:
p
(
x
) = 4
x
3 + 8
x
2 – 11
x
– 15 = (
x
+ 1)(2
x
+ 5)(2
x
– 3) 26 of 27 © Boardworks Ltd 2005
Examination-style question
c) The graph of
y
=
p
(
x
) crosses the
x
-axis when 4
x
3 + 8
x
2 – 11
x
– 15 = 0 (
x
+ 1)(2
x
+ 5)(2
x
– 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
y
Also the coefficient of
x
3 is positive so the graph is -shaped: –2.5
–1 –15 1.5
x
27 of 27 © Boardworks Ltd 2005