lOMoARcPSD|36962535
(1.1) Operations with polynomials
cubic
Li
32:asc
in
polynomial:
of
In
e.g.
polynomial
-all
variables
(..1.1)
Addition
Example
1:
raised
are
(5x24
3x
-
with
term
(7x"
3x
- 2x
5x
+
+3x
-
2)
2
-
(x3
another
one
+3x
x5
terms
=
add
multiplying,
When
*
the
exponents
by
dividing
-
2)(x2
2x4
-
65
=
-
-
4x
2"
-
2x
-
4)
-
3x3
+
x3
6x2
-
8
-
-
-
12x
4x
+
8
+
8
8
-
222
+
Division
c.1.3)
3
Example
-
3x2
Method
+
x
-
6)
(x
=
2
2
2x3
=
Note:
if
e.g.
leave
3
Write
x23
NOW
2
sc
/firstterm
in
2)
2x(x
-
2)
space.
5
+2x
+
subtractec-4x"
-
by
i)
5as
2x
+
2x3
in
+
+
2x2
<firstterm
missing
is
blank
a
2x
6
-
<
dividend
the
term,
x
+
4x"
-
found
I
3x
-
2x3
a
2)
-
I
2x
-
x
or
to
+
2x
I
positive power
2x)
+
contains
it
"3"
order
52
a
3
resulting
add
x
-
variable
Example 2
-multiply each
-
to
<polynomial)
a
of
Multiplication
(1.1.2)
x
is
notcontain
does
Add
(x3)
cubic
a
=12x"
(2x
&
+
highestpower
order
a
bost(x
+
x
+
3x
->
x
-
+
x
2x3-3x
from
x-6
+
and
bring
down
the
nextterm
:x
method
6
↓
N
-
2
-
2
-
if
know
you
there's
remainder
no
x2 x
+
x
x(x
-
2x
+
-
(2x3
Let
2)
sides
x
both
2x3
continuing gives:
2x2
x
-
2
2x3
-
x( 3)
+
+
3x
x
-
+
multiply
x2 x
+
+2x
2
-
-
-
x
-
+
x
+
-
coefficients
a
=
3 b
=
-
2a
b 1
=
362
-
6
2
=
3x2
-
1
-
6
c
0
:(2x
-
3x
x
+
-
6)
(x
=
-
2)
2x2
=
x
+
3
+
..
b)
=
b
=
ax
(x 2)
=
-
bx
2
+
+
(x-2)
32
3x2
-
compare
-
-
by
x
+
(ax
c)(x 2)
bx
+
+
-
out
2x3
6
1
n
-
x
quotient
3x2
-
-
26
3
=
2x2
+x
3
+
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b
ax"
=
(b
+
-
za)x2
(c 2b)x
+
-
-
2x
lOMoARcPSD|36962535
(1.2) solutions of polynomial equations
formula.
Quadratic
x =
-
Ib2
b
2
4ac
-
-
solution: ax +bx
c
If
(x
k
to
-
a)
is
is
is
the
of
of
factor
a
lax-b)
Example
cubic:
roots
finding
when
the
graph
the expression
of
outs
x-axis
the
=
factor
a
lax-b)
then
a
method
numerical
fix),
the
polynomial
then
f(a)
0
=
and
s= a
is
a
root
equation
the
of
factor
of
of
factor
f(x) x3
=
show
(ii)
Solve
itis
necessary
23
=
x2
the
-
22
-
f(x)
eq.
f(),
then
+(a)
0
=
and
x =
a
is
a
rootof
f(x)
3x2
2
3x
+
factor
a
=
0
f(2)
that
show
to
-
is
that(x-2)
(i)
f(z)
-
0
=
2
+
=0
therefore
(ii)
is
it
(x-2)
Since
is
(x
-
factor,
a
+x
x
factor
a
divide
you
f(x)
by it
1
=
-
2)xP
x
2
-
-
x
-
3x
2x2
12
2
+
d
-
3x
- 2
2
-x
-
x
+
2
+
0
so
f(z) 0
=
becomes
x2
-1.618
=
(x
-
2)(x
or
f(x) 0.
=
if
f(a)
conversely,
0,
=
then
f(x):
a
a
polynomial
I
that
Given
(i)
is
if
so
more
or
a
theorem
a)
-
spotting,
of
0
+
-
factor
solution
a
~
The
for
methods
2
0.618
x
1) 0
=
+
-
or
2
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the
equation
f(x) 0.
=
conversely,
f(a) 0
if
=
lOMoARcPSD|36962535
(1.3) the modulus function
modulus
(e.g.)
↳
absolute
his
magnitude
131
value
without
the
and
1-31
if
s2
O
3
=
1x1
E
=
x
If
2
-
sigh
3
modulus
->
=
cannotbe
x <O
examples
rules
(c)
|a
1
b)
1b
1
2
1b)
1
=
acx
(a
a
x<
a
a|<b
a
9
>
-
b2
A
-
=
K
statement: (SL)
18
=
1x1
k
where
=
1
131
a)
-
x
-
(x)
the
-
=
12
(a)
1x
x|
=
-
1x
negative
-
or
(
=
1x |
>
b
1x
k
Or
=
x
-
+
3
=
3)2
13)
x>a
+
8)
-
3
-
3
1- 32
=
=
..
?
0
15
31
31
31
=
1x1
b<x<a
-
51
-
-
-
-
=
-
3
3[x
x
2175
-
=
-
=
3
3
>
x
or
3
3(x(7
k
=
-
↓
consider
b)
(ax
k
+
lax
b)
=
d
ax+b
+
;
=
+
cx
+
k
ax b
=
ax
=
ax
b
-
+
=
d;ax b
+
+
=
I
k
-
(xx
d)
+
12x
2x
-
-
2x
1x
=
3
i
3
=
2x
L
2
=
2
x
-
1
+
2x
=
x
12
1
11
x
1)
back
if
in
12x
3
=
-
3
=-
2
=
=-
-
x
-
-
2
-
2x
=
2x
=
4
x
=
x
1
11
3
1
check
=
-
5
-
1-
41
91
see
I
h)
-
to
work
they
example
Example
putresults
always
=
-
i
1
+
-
5 1
+
1
09
a
a
-
b*
write
-
=(a)
191
=
1(x b))
+
1(x f))
=
+
use
the
if
191
>b
if
191
<b
rule
|a
as
1b))
+
b2
a =
(b)
=
1a1=1b1
Live
1b12
-
1b1))191
-
E
a
-
a<
<(
=
a
b2
ifb 0
=
b2
b2
=
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+
-
-
1
=
(2x 1)
=
2x
x =
wrong.
=
4
5
modulus
al
-
=-
↓
I
x
5
=
2x
1
+
41
-
31
-
1
I
2x1
=
3
=
+ 1
·x
1
=
lOMoARcPSD|36962535
example
example
n)
13x
5
=x
+
2x
3x2
i
+
4
+
4x
I
-
x
(a)
+
=
4
3x
5/
1x
+
x
-
-
-
5
13x
x
1b)
4)
(x
=
+
- 14
qx2 + 24x
8x2 +
14x
13x2
check:
(x 5)
31
using
10
=
3
x
x
-10
+
1x
-
9
+
25
+
0
=
1)(4x 9)
0
=
+
51
+
5)(1);
12
-
x
+
3
+
(
=
x
+
5
2x
x
1
=
5)
(x
+
= 7
+
51
+
-
10(2)
using
eq
..2
(2x
+
-
10x
=
10
+
1x +
31
7
n1W
4)
5)2
=
+
+
-
+
=
-
Ex
=
2
example
1x
5)/13)
12
4)
+
-
x
+
x2
16
-
5)
(x
+
=
b2
+
=
+
=
a
=
(2x 4)2
9
9
I
-
-
=
method
alt
I
+
-
7
=
-
x
2
x
x
x
5
+
0
=
=
-
=
-
(7
-12
-
x)
C.:
3
1
x
=
-
=
+
1 51
+
false solution)
x
12 51
+
x
=
5
+
⑧
=
or
-
eq 2
=
=
x
g
-
10
13
+
13
+
x
(false)
9
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5
+
x
=
-
=
13)
x
-
+
q
=
/4
lOMoARcPSD|36962535
(1.3.1) Graphs of modulus functions
steps:1.
draw
y=
x
↑
c axis
in
reflext
2
=
y x
example
(x)
=
y
I
sketch
Ex-1).
1
=
y
alternative
show
points
the
where
the
meets
graph
the
axes.
Use
the
graph
/E-11
express
to
form
-firstsketch
y
Ex-1
=
(1)
I
y
↑...
0
=
2x 1
=
x
x
reflectin
(tx
-
2
=
0;y
=
↳
y
Ex
=
-x
-
(Ex 1)
if
y
12x
=
11
3
!
I
found
2
roots
at
=-1
and
x
=
!
-
-
1
y
-
I
12x
=
2
11 3
=
-
3
41
LD
1)
=
We
example
-
x<2
=
-
1
1ifx22
-
y 2x
1x
-
2
example
-
1
=
·-
12x
-
=
s-axis
the
1)
(2)
2x
=
root:
1
+
2
I
=
↑
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i
I
H
y
3
=
in
an
lOMoARcPSD|36962535
(1.3.2) Solving modulus inequalities
main
rules:
(2)
-
b
1191
=
191
b
b =a =
a
Ib
<
-
b
or
a
b
example I
<3
12x-51
solve
⑦
⑧
·
↳
y
12x
=
-
51
2x
-
<
3
2x
-
52 3
2
<Ix
-8
1
a
-
-
4x22
-
4x
9
=
1
+
-
1912
<>
(b)
=
-
3x2
-
1
x
8
x)(x 2)
+
:.x(
2x
-
-
5fs)
(2x
-
bx
-
=
2.5
5)fx
(2.5
-
x
=
=
-
-
2
3
-
1)
2x
-
1)
=
-
- (x
=
3x
x
-
fx z
1
fx2t
3-x
fxx?3
-
3
13
-
x)
=
-
43
2x
-
(2x 1)
·B
2
0
-
13x
=
12x
0
2;x
= b2
y
2
x2
+
=
=
a
x)2
1
(3x
=
↳
1) 2(3
+
2x
-
51
-
2.5
- 11213 -x1
(2x
12x
2
example
12x2
y
3
=
3)
4
=
4/3
=
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(3 x)
-
x
+
3
lOMoARcPSD|36962535
(2.1) Exponential functions
with
graphs
all
is
-x-axis
-increase
y
zreflect
->
=
with
an
gradient
increasing
rate
3-
=
y-axis
the
in
y z.
(0,1)
through
growth
iy
=
pass
asymptote
ever-increasing
atan
will
y=
horizontal
a
exponential
->
ka"
form
a
z"
=
y
A
decay
exponential
(2.2) Logarithms
general
rule:
y
logarithms
a:
=
base
the
to
logay
10
number
any
positive
e.g.
109101000:
x
=
be
can
3
10910100
log,010
expressed
109,01
2
through
base
0
109,0 (4) 109,0(10-3
=
=
1
=
=
-1
=
(160) 10910(10-3)
log,0
10
=
=
-2
logarithms
The
laws
(1)
multiplication
of
9109a
=
alogacy
RootS
(5)
x
glogaY
109
10gx
x
=
log"
:109accy 10gax 10gay
=
+
loga
-
division
(2)
(6)
109, (f)
10ga"
=
-
the
logarithm
a
of
number
to
its
own
base
10gay
since
·109,
(3)
-logs
6
5' =
5,
10955=1
1
=
indices
x
=
nxn....
(7)
Reciprocals
logs loge logx....
+
=
log
..
(4)
logsch nlogx
loga (8)
=
Power
(5)
=
-logy
10gac -logay
=
log (5)
log1-logy
=
zero
I
0-1094
=-
as
a
1;10gal
=
logy
0
=
-any
base
>I
=- logy
between
is
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used
o
and
I
is
negative
lOMoARcPSD|36962535
(2.3) Graphs of logarithms
whatever
the
value,
of
as
(a21),
base
the
the
graph
y-logab
has
of
the
same
general
shape
yu
...
properties:
--
- axis: (1,0)
I
i
a
I
0
values
only
pos
x
asymptote
0:
=
0xx>1:
-
no
negative
for
lim
12
of
heightbutgrad consistently
·(a,1)
y 10gak
=
109, (9")
-
x
x
and
=
=
al
loga"
=(
d
(2.4) Modelling curves
rc"
y
=
or
y
curve:
line:
straight
nlogd
logR
t
=
y
ux
n
)
<=
=
=>
K 1
intercept
=
d
M3
=
y
109,0k=intercept
logt
N
↓
.
ka
=
C
I
logy logh
=
nlogx
+
:y
a
=
=)
logy-logh
scloga
+
(2.5) The natural logarithm
y
z
=
=>
Ina-lub:
·
·In
as
1
=
In
Inb
Ina
Si Ins
(n(ab)
+
=
0
=
Inx-
x-0
-
&
(2.6) The exponential function
y
e
Inc
=
is
=x
the
e
=
inverse
In
of
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decreases
lOMoARcPSD|36962535
(3.1) Reciprocal trigonometry functions
important
coses
-
formulae
sn
·cosecO
->
=
SecO
undefined
sind
=
0
for
at
iiiIt !
I
i
I
I
I
I
I
I
!
y 20Se
i
y + AME
=
I
identities
cos28=I
Sin2 8 +
Sin O
(ii)
=
tan 28
example
1
+
SinZO
Sin &
sec
c
-
=>
points
!
=
cos2
those
=
(sin8
-an
i
(i)
8 180,360
⑦
=
·OtO
is
+
t
sin? &
Sec2O
=-
=
cos2O=
cos"O
Sin
+
cOt20
I
I
=
sin28
coses? &
=
i
I
CoSe2120
=
sin 120
2
I
5
(3.2) Compound-angle formulae
sin(x
y)
sin(x
cos
sinosy
y)
-
-
coscosy
+
cos(x
y)
-
tansL
an(x
*
-
y)
It
-
-
tany
tansctany
cost
cos(-0)
Sin
+
tany
+
-tanxtany
=
+
sinising
sinising
coscosy
y)
(x
cossing
-
-
=
tansz
tan
+
=sincosy
y)
(x
cosssing
=
+
=
(-0)
=
-sinQ
(3.3) Double angle formulae
SinzO=2 sinACOSO
sin
·cost
cos28
(1-cOS28S
=
E (1 cos28)
+
=
cos28-sin28
=
COS28=1
-
2 Sin28
COSL8 =2cOS"8-1
It an O
tan20=
-tan 20
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lOMoARcPSD|36962535
(3.4) The forms of (theta alpha), rsin(theta alpha)
I
I
-
-asino
-
bcost
+
..
c
=
rsin(8 +a) r
=
asint
-r
(sinAcosa
bcost=rsin (0
+
a
=
a)
+
+
cosAsina)
Il
a
cost
bsint
+
rcos(8
=
a)
+
b2
+
=
-sina
r
-cosa
A
=
r
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