A-level Mathematics/CIE/Pure Mathematics 2/Algebra

The Modulus Function

The modulus function $|x|$ returns the magnitude of $x$. For instance, $|-3|$ will return $3$, $|5|$ will return $5$.

The modulus function can be defined as $|x|=\{\begin{array}{cc}-x,& \text{if }x<0\\ x,& \text{if }x\ge 0\end{array}$.

The modulus function as a graph.

Graphing the modulus function

Graphs of the modulus function are just a straight-line graph that has been reflected for negative output values. The graph of $y=|ax+b|$ is like the graph of $y=ax+b$ except that every point below the x-axis folds upwards to produce a V-shaped graph.

Here is an interactive graph which shows the relationship between the graph of a line and the graph of the modulus of that line.

Solving Equations & Inequalities

To solve equations and inequalities involving the modulus function, we can square both sides.

e.g. Solve $|4x+2|>|2x-3|$

$\begin{array}{cc}|4x+2|& >|2x-3|\\ (4x+2{)}^2& >(2x-3{)}^2\\ 16x^2+16x+4& >4x^2-12x+9\\ 12x^2+28x-5& >0\\ x^2+\frac{7}{3}x& >\frac{5}{12}\\ (x+\frac{7}{6}{)}^2& >\frac{15}{36}+\frac{49}{36}\\ x+\frac{7}{6}>\sqrt{\frac{64}{36}}& \text{ or }x+\frac{7}{6}<-\sqrt{\frac{64}{36}}\\ x+\frac{7}{6}>\frac{8}{6}& \text{ or }x+\frac{7}{6}<-\frac{8}{6}\\ x>\frac{1}{6}& \text{ or }x<-\frac{15}{6}\\ \text{In interval notation, }x& \in (-\mathrm{\infty },-\frac{5}{2})\cup (\frac{1}{6},\mathrm{\infty })\end{array}$

An alternative method is to look at the places where the functions inside the modulus change sign, i.e. where $f(x)=0$.

$4x+2$ changes sign at $x=-\frac{1}{2}$

$2x-3$ changes sign at $x=\frac{3}{2}$

$\begin{array}{ccc}\text{Where }x<-\frac{1}{2}& \text{Where }-\frac{1}{2}\le x<\frac{3}{2}& \text{Where }\frac{3}{2}\le x\\ -(4x+2)>-(2x-3)& 4x+2>-(2x-3)& 4x+2>2x-3\\ 4x+2<2x-3& 4x+2>-2x+3& 4x+2>2x-3\\ 2x<-5& 6x>1& 2x>-5\\ x<-\frac{5}{2}& x>\frac{1}{6}& x>-\frac{5}{2}\\ & & \text{Out of range}\end{array}$

$\text{In interval notation, }x\in (-\mathrm{\infty },-\frac{5}{2})\cup (\frac{1}{6},\mathrm{\infty })$

Dividing Polynomials

Dividing polynomials uses the same method as dividing numbers with long division.

Dividing Numbers

Suppose we need to find $22253÷17$. We can use the method of long division:

  ______
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, $22253÷17=1309$

Dividing Polynomials

We can use the same method to divide polynomials.

e.g. $(x^3+2x^2+2x+1)÷(x+1)$

      ____________________
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, $(x^3+2x^2+2x+1)÷(x+1)=x^2+x+1$

The Remainder Theorem

A remainder occurs when the divisor does not fit into the dividend a whole number of times.

e.g. $22256÷17=\frac{22253}{17}+\frac{3}{17}=1309+\frac{3}{17}$ has a remainder of $3$.

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 $1$.

This can be expressed as $x^3+3x^2+3x+3=(x^2+x+1)(x+2)+1$

In general, a quotient and remainder can be expressed as $Dividend=(Quotient)(Divisor)+Remainder$

This expression leads to a useful theorem in mathematics: the remainder theorem.

If we divide a polynomial $p(x)$ by a given divisor $(ax-b)$, the expression can be written as $p(x)=(Quotient)(ax-b)+Remainder$.

If we substitute the value of $b/a$ into the polynomial, we get:

$\begin{array}{cc}p(b/a)& =(Quotient)(a(b/a)-b)+Remainder\\ p(b/a)& =(Quotient)(b-b)+Remainder\\ p(b/a)& =(Quotient)(0)+Remainder\\ p(b/a)& =Remainder\\ \end{array}$

Thus, the remainder theorem states that for a given polynomial $p(x)$, $p(b/a)$ gives the remainder obtained from $\frac{p(x)}{ax-b}$.

e.g. If $p(x)=x^3+3x^2+3x+3$, $p(-2)$ will give the remainder obtained from $(x^3+3x^2+3x+3)÷(x+2)$:

$\begin{array}{cc}p(-2)& =(-2{)}^3+3(-2{)}^2+3(-2)+3\\ & =-8+3(4)-6+3\\ & =-8+12-3\\ & =1\end{array}$

The Factor Theorem

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 $p(b/a)=0$, $(ax-b)$ is a factor of $p(x)$

e.g. $p(x)=x^3-5x^2+x+10$. Use the factor theorem to find a factor of $p(x)$.

$\begin{array}{ccc}p(1)& =1^3-5(1^2)+1+10=1-5+1+10=7& \ne 0\\ p(-1)& =(-1{)}^3-5(-1{)}^2-1+10=-1-5-1-10=-17& \ne 0\\ p(2)& =2^3-5(2^2)+2+10=8-5(4)+2+10=8-20+12& =0\\ \therefore & (x-2)\text{ is a factor of }{x}^3-5x^2+x+10\end{array}$

Notes