The Modulus Function
edit
The modulus function [ note 1]
|
x
|
{\displaystyle |x|}
x
{\displaystyle x}
|
−
3
|
{\displaystyle |\!-\!3|}
3
{\displaystyle 3}
|
5
|
{\displaystyle |5|}
5
{\displaystyle 5}
The modulus function can be defined as
|
x
|
=
{
−
x
,
if
x
<
0
x
,
if
x
≥
0
{\displaystyle |x|={\begin{cases}-x,&{\text{if }}x<0\\x,&{\text{if }}x\geq 0\end{cases}}}
The modulus function as a graph. Graphing the modulus function
edit
Graphs of the modulus function are just a straight-line graph that has been reflected for negative output values. The graph of
y
=
|
a
x
+
b
|
{\displaystyle y=|ax+b|}
y
=
a
x
+
b
{\displaystyle y=ax+b}
Here is an interactive graph which shows the relationship between the graph of a line and the graph of the modulus of that line.
edit
To solve equations and inequalities involving the modulus function, we can square both sides.
e.g. Solve
|
4
x
+
2
|
>
|
2
x
−
3
|
{\displaystyle |4x+2|>|2x-3|}
|
4
x
+
2
|
>
|
2
x
−
3
|
(
4
x
+
2
)
2
>
(
2
x
−
3
)
2
16
x
2
+
16
x
+
4
>
4
x
2
−
12
x
+
9
12
x
2
+
28
x
−
5
>
0
x
2
+
7
3
x
>
5
12
(
x
+
7
6
)
2
>
15
36
+
49
36
x
+
7
6
>
64
36
or
x
+
7
6
<
−
64
36
x
+
7
6
>
8
6
or
x
+
7
6
<
−
8
6
x
>
1
6
or
x
<
−
15
6
In interval notation,
x
∈
(
−
∞
,
−
5
2
)
∪
(
1
6
,
∞
)
{\displaystyle {\begin{aligned}|4x+2|&>|2x-3|\\(4x+2)^{2}&>(2x-3)^{2}\\16x^{2}+16x+4&>4x^{2}-12x+9\\12x^{2}+28x-5&>0\\x^{2}+{\frac {7}{3}}x&>{\frac {5}{12}}\\(x+{\frac {7}{6}})^{2}&>{\frac {15}{36}}+{\frac {49}{36}}\\x+{\frac {7}{6}}>{\sqrt {\frac {64}{36}}}&{\text{ or }}x+{\frac {7}{6}}<-{\sqrt {\frac {64}{36}}}\\x+{\frac {7}{6}}>{\frac {8}{6}}&{\text{ or }}x+{\frac {7}{6}}<-{\frac {8}{6}}\\x>{\frac {1}{6}}&{\text{ or }}x<-{\frac {15}{6}}\\{\text{In interval notation, }}x&\in \left(-\infty ,-{\tfrac {5}{2}}\right)\cup \left({\tfrac {1}{6}},\infty \right)\end{aligned}}}
An alternative method is to look at the places where the functions inside the modulus change sign, i.e. where
f
(
x
)
=
0
{\displaystyle f(x)=0}
4
x
+
2
{\displaystyle 4x+2}
x
=
−
1
2
{\displaystyle x=-{\frac {1}{2}}}
2
x
−
3
{\displaystyle 2x-3}
x
=
3
2
{\displaystyle x={\frac {3}{2}}}
Where
x
<
−
1
2
Where
−
1
2
≤
x
<
3
2
Where
3
2
≤
x
−
(
4
x
+
2
)
>
−
(
2
x
−
3
)
4
x
+
2
>
−
(
2
x
−
3
)
4
x
+
2
>
2
x
−
3
4
x
+
2
<
2
x
−
3
4
x
+
2
>
−
2
x
+
3
4
x
+
2
>
2
x
−
3
2
x
<
−
5
6
x
>
1
2
x
>
−
5
x
<
−
5
2
x
>
1
6
x
>
−
5
2
Out of range
{\displaystyle {\begin{array}{ccc}{\text{Where }}x<-{\frac {1}{2}}&{\text{Where }}-\!{\frac {1}{2}}\leq x<{\frac {3}{2}}&{\text{Where }}{\frac {3}{2}}\leq x\\-(4x+2)>-(2x-3)&4x+2>-(2x-3)&4x+2>2x-3\\4x+2<2x-3&4x+2>-2x+3&4x+2>2x-3\\2x<-5&6x>1&2x>-5\\x<-{\frac {5}{2}}&x>{\frac {1}{6}}&x>-{\frac {5}{2}}\\&&{\text{Out of range}}\\\end{array}}}
In interval notation,
x
∈
(
−
∞
,
−
5
2
)
∪
(
1
6
,
∞
)
{\displaystyle {\text{In interval notation, }}x\in \left(-\infty ,-{\tfrac {5}{2}}\right)\cup \left({\tfrac {1}{6}},\infty \right)}
Dividing Polynomials
edit
Dividing polynomials uses the same method as dividing numbers with long division.
To do:
Suppose we need to find
22253
÷
17
{\displaystyle 22253\div 17}
______
17|22253
__1___
17|22253 17 goes into 22 once with 5 left over
-17↓
52 Next we bring down the 2
__13__
17|22253
-17↓↓
52↓ 17 goes into 52 thrice with 1 left over
-51↓
15 Next we bring down the 5
__1309
17|22253
-17↓↓↓ 17 doesn't go into 15, so we bring down the 3
52↓↓
-51↓↓
153 17 goes into 153 nine times with nothing left over
-153
0
Thus,
22253
÷
17
=
1309
{\displaystyle 22253\div 17=1309}
Dividing Polynomials
edit
We can use the same method to divide polynomials.
e.g.
(
x
3
+
2
x
2
+
2
x
+
1
)
÷
(
x
+
1
)
{\displaystyle (x^{3}+2x^{2}+2x+1)\div (x+1)}
____________________
x + 1 |x^3 + 2x^2 + 2x + 1
________x^2_________
x + 1 |x^3 + 2x^2 + 2x + 1 (x + 1) goes into (x^3 + 2x^2) x^2 times with x^2 left over
-(x^3 + x^2) ↓
x^2 + 2x Bring down the 2x
________x^2_+__x____
x + 1 |x^3 + 2x^2 + 2x + 1 (x + 1) goes into (x^2 + 2x) x times with x left over
-(x^3 + x^2) ↓ ↓
x^2 + 2x ↓ Bring down the 1
-(x^2 + x) ↓
x + 1
________x^2_+__x___1
x + 1 |x^3 + 2x^2 + 2x + 1 (x + 1) goes into (x + 1) once with nothing left over
-(x^3 + x^2) ↓ ↓
x^2 + 2x ↓
-(x^2 + x) ↓
x + 1
-(x + 1)
0
Thus,
(
x
3
+
2
x
2
+
2
x
+
1
)
÷
(
x
+
1
)
=
x
2
+
x
+
1
{\displaystyle (x^{3}+2x^{2}+2x+1)\div (x+1)=x^{2}+x+1}
The Remainder Theorem
edit
A remainder occurs when the divisor does not fit into the dividend a whole number of times.
e.g.
22256
÷
17
=
22253
17
+
3
17
=
1309
+
3
17
{\displaystyle 22256\div 17={\frac {22253}{17}}+{\frac {3}{17}}=1309+{\frac {3}{17}}}
3
{\displaystyle 3}
It can also occur in polynomials:
____________________
x + 2 |x^3 + 3x^2 + 3x + 3
________x^2_________
x + 2 |x^3 + 3x^2 + 3x + 3
-(x^3 + 2x^2) ↓
x^2 + 3x
________x^2_+__x____
x + 2 |x^3 + 3x^2 + 3x + 3
-(x^3 + 2x^2) ↓ ↓
x^2 + 3x ↓
-(x^2 + 2x) ↓
x + 3
________x^2_+__x_+_1
x + 2 |x^3 + 3x^2 + 3x + 3
-(x^3 + 2x^2) ↓ ↓
x^2 + 3x ↓
-(x^2 + 2x) ↓
x + 3
-(x + 2)
1
Here, the remainder is
1
{\displaystyle 1}
This can be expressed as
x
3
+
3
x
2
+
3
x
+
3
=
(
x
2
+
x
+
1
)
(
x
+
2
)
+
1
{\displaystyle x^{3}+3x^{2}+3x+3=(x^{2}+x+1)(x+2)+1}
In general, a quotient and remainder can be expressed as
D
i
v
i
d
e
n
d
=
(
Q
u
o
t
i
e
n
t
)
(
D
i
v
i
s
o
r
)
+
R
e
m
a
i
n
d
e
r
{\displaystyle Dividend=(Quotient)(Divisor)+Remainder}
This expression leads to a useful theorem in mathematics: the remainder theorem.
If we divide a polynomial
p
(
x
)
{\displaystyle p(x)}
(
a
x
−
b
)
{\displaystyle (ax-b)}
p
(
x
)
=
(
Q
u
o
t
i
e
n
t
)
(
a
x
−
b
)
+
R
e
m
a
i
n
d
e
r
{\displaystyle p(x)=(Quotient)(ax-b)+Remainder}
If we substitute the value of
b
/
a
{\displaystyle b/a}
p
(
b
/
a
)
=
(
Q
u
o
t
i
e
n
t
)
(
a
(
b
/
a
)
−
b
)
+
R
e
m
a
i
n
d
e
r
p
(
b
/
a
)
=
(
Q
u
o
t
i
e
n
t
)
(
b
−
b
)
+
R
e
m
a
i
n
d
e
r
p
(
b
/
a
)
=
(
Q
u
o
t
i
e
n
t
)
(
0
)
+
R
e
m
a
i
n
d
e
r
p
(
b
/
a
)
=
R
e
m
a
i
n
d
e
r
{\displaystyle {\begin{aligned}p(b/a)&=(Quotient)(a(b/a)-b)+Remainder\\p(b/a)&=(Quotient)(b-b)+Remainder\\p(b/a)&=(Quotient)(0)+Remainder\\p(b/a)&=Remainder\\\end{aligned}}}
Thus, the remainder theorem states that for a given polynomial
p
(
x
)
{\displaystyle p(x)}
p
(
b
/
a
)
{\displaystyle p(b/a)}
p
(
x
)
a
x
−
b
{\displaystyle {\frac {p(x)}{ax-b}}}
e.g. If
p
(
x
)
=
x
3
+
3
x
2
+
3
x
+
3
{\displaystyle p(x)=x^{3}+3x^{2}+3x+3}
p
(
−
2
)
{\displaystyle p(-2)}
(
x
3
+
3
x
2
+
3
x
+
3
)
÷
(
x
+
2
)
{\displaystyle (x^{3}+3x^{2}+3x+3)\div (x+2)}
p
(
−
2
)
=
(
−
2
)
3
+
3
(
−
2
)
2
+
3
(
−
2
)
+
3
=
−
8
+
3
(
4
)
−
6
+
3
=
−
8
+
12
−
3
=
1
{\displaystyle {\begin{aligned}p(-2)&=(-2)^{3}+3(-2)^{2}+3(-2)+3\\&=-8+3(4)-6+3\\&=-8+12-3\\&=1\end{aligned}}}
The factor theorem is a special case of the remainder theorem for when the remainder is zero.
If the remainder is zero, that means that the divisor is a factor of the dividend.
Thus, if
p
(
b
/
a
)
=
0
{\displaystyle p(b/a)=0}
(
a
x
−
b
)
{\displaystyle (ax-b)}
p
(
x
)
{\displaystyle p(x)}
e.g.
p
(
x
)
=
x
3
−
5
x
2
+
x
+
10
{\displaystyle p(x)=x^{3}-5x^{2}+x+10}
p
(
x
)
{\displaystyle p(x)}
p
(
1
)
=
1
3
−
5
(
1
2
)
+
1
+
10
=
1
−
5
+
1
+
10
=
7
≠
0
p
(
−
1
)
=
(
−
1
)
3
−
5
(
−
1
)
2
−
1
+
10
=
−
1
−
5
−
1
−
10
=
−
17
≠
0
p
(
2
)
=
2
3
−
5
(
2
2
)
+
2
+
10
=
8
−
5
(
4
)
+
2
+
10
=
8
−
20
+
12
=
0
∴
(
x
−
2
)
is a factor of
x
3
−
5
x
2
+
x
+
10
{\displaystyle {\begin{aligned}p(1)&=1^{3}-5(1^{2})+1+10=1-5+1+10=7&\neq 0\\p(-1)&=(-1)^{3}-5(-1)^{2}-1+10=-1-5-1-10=-17&\neq 0\\p(2)&=2^{3}-5(2^{2})+2+10=8-5(4)+2+10=8-20+12&=0\\\therefore \ &(x-2){\text{ is a factor of }}x^{3}-5x^{2}+x+10\end{aligned}}}
Notes
↑ Also known as the absolute value function
Logarithmic and Exponential Functions →
