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 $$|x| = \begin{cases}-x,& \text{if } x < 0 \\ x,& \text{if } x\geq 0 \end{cases}$$.



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 $$|4x+2| > |2x-3|$$

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

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 $$x=\frac{3}{2}$$

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

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

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

Dividing Numbers
Suppose we need to find $$22253 \div 17$$. 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, $$22253 \div 17 = 1309$$

Dividing Polynomials
We can use the same method to divide polynomials.

e.g. $$(x^3 + 2x^2 + 2x + 1) \div (x + 1)$$

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

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

e.g. $$22256 \div 17 = \frac{22253}{17} + \frac{3}{17} = 1309 + \frac{3}{17}$$ 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:

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

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 $$(x^3 + 3x^2 + 3x + 3) \div (x + 2)$$:

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

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 $$p(x)$$.

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


 * Notes