Polynomials Part1

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Notes

On the surface, factor theorem & remainder theorem seems easy when compared to other subtopics. However, a lot of student make the mistake and think that this subtopic is about CALCULATION. It is NOT (calculator will handle most of the calculations anyway). It is about THEOREM and how to USE it CORRECTLY. A correct calculation with the wrong way of using the theorem will result in LESS/ZERO marks. There will also be a lot of WRITING (of words, not just numbers and symbols), which is another thing student find it hard to do. Failure to learn the usage of the relatively simple theorems here will also probably mean failure to grasp many other subtopics in the syllabus.

Learning Objectives (Syllabus)

understand the meaning of the degrees and coefficients of polynomials
carry out elementary operations on polynomials
use the condition for the equality of two polynomials
find the factors and zeroes of polynomials
prove and use the remainder and factor theorems

Prior Knowledge

Basic algebraic skills

Degree & Coefficients

$P\left (x \right )=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\dots+a_{n-1}x+a_{n}$ where $\,n$ is a positive integer.

degree :
terms : $a_{0}x^{n},a_{1}x^{n-1},\dots$
coefficients : $a_{0},a_{1},\dots$

Example

$P\left (x \right )=4x^{3}-x^{2}+7$

$P\left (x \right )$ is a polynomial of degree $\,3$
How do we refer to the and ?
- Coefficient of $\,x^{3}$ (of the term in $\,x^{3})$ $\,=4$
- Term in $\,x^{3}$ $\,=4x^{3}$
Simple enough, but MAKE SURE you don't give term when asked for coefficient or vice versa.
- Coefficient of $\,x^{2}$ $\,=-1$
- Term in $\,x^{2}$ $\,=-x^{2}$
- Coefficient of $\,x$ Huh? Where got $\,x$ ? $\,=0$
- How about the $\,7$ ?

Common Names

linear - degree 1 Ex: $P\left( x \right)=3x-1$
quadratic - degree 2 Ex: $P\left( x\right)=5x^{2}+3x+2$
cubic - degree 3 Ex: $P\left(x \right)= -2x^{3}-x^{2}-1$

Function vs Equation

Polynomials is a type of function and NOT equation.

Example

$f\left (x \right )=x^{2}-1$ is a

$\,x^{2}-1=0$ is a

Value of Polynomial

The value of a polynomial $P\left(x \right)$ when $\,x=a$

is written as $P\left(a \right)$
and is calculated by substituting $\,x=a$ into the polynomial
NOTE:

Example: $P\left(x \right)=4x^{3}-x^{2}+7$

$P\left(1\right)$ $\,=10$
$P\left(-2\right)$ $\,=-29$
$P\left(\frac{1}{2}\right)$ $=\frac{29}{4}$
$P\left(-\frac{2}{3}\right)$ $=\frac{145}{27}$
$P\left(a\right)$ $\,=4a^{3}-a^{2}+7$
$P\left(-a\right)$ $\,=-4a^{3}-a^{2}+7$
$P\left(2a\right)$ $\,=32a^{3}-4a^{2}+7$
$P\left(\frac{1}{2}a\right)$ $=\frac{1}{2}a^{3}-\frac{1}{4}a^{2}+7\,$

Algebraic Operation

We can apply addition, subtraction and multiplication to polynomials by using basic algebra laws.

Example: $P\left(x \right)=x^{2}+1, Q\left(x \right)=2x+1$

$P\left(x \right)+Q\left(x \right)$ $\,=x^{2}+2x+2$
$P\left(x \right)-Q\left(x \right)$ $\,=x^{2}-2x$
$P\left(x \right)\cdot Q\left(x \right)$ $=\left(x^{2}+1 \right)\left(2x+1 \right)$ $\,=2x^{3}+x^{2}+2x+1$

Note : If we multiply polynomial of degree $\,m$ with polynomial of degree $\,n$ , we get polynomial of degree $\,m+n$

Division of Polynomial

This is where the fun begins. :P Before we look at division of polynomials, lets revise division of numbers. Yes, the method (known as long division) that we learn in primary school

$17 \div 5 = ?$
$\begin{array}{rl} & 5\overline{)17} \\ \end{array}$ $\begin{array}{rl} & \quad \;3 \\ & 5\overline{)17} \\ \end{array}$ $\begin{array}{rl} & \quad \;3 \\ & 5\overline{)17} \\ & \quad\underline{15} \\ \end{array}$
$\begin{array}{rl} & \quad \;\;\;3 \\ & \;\;5\overline{)17} \\ & {\color{Red}\left(-\right)}\underline{15} \\ & \qquad 2 \\ \end{array}$
Thus, $\frac{17}{5}=$ $\,3$ $+\frac{2}{5}$
We can also rearrange it to $\,17=$ $5\times 3 +2$
Once again, from $\begin{array}{rl} & \quad \;{\color{Blue}3} \\ & {\color{Red}5}\overline{)17} \\ & \quad\underline{15} \\ & \quad\;\; {\color{Green}2} \\ \end{array}$ $\frac{17}{{\color{Red}5}}={\color{Blue}3}+\frac{{\color{Green}2}}{{\color{Red}5}}$ or $17 = {\color{Red}5} \times {\color{Blue}3} +{\color{Green}2}$
Let's revise the terms used to describe these numbers
- The ${\color{Red}5}$ is the divisor
- The ${\color{Blue}3}$ is the quotient
- The ${\color{Green}2}$ is the remainder

Example

$\left(x^{2}+7x-5\right) \div \left( x + 2\right)$
The one thing to keep in mind is that the steps are the same as when it involves numbers
- $\begin{array}{rl} & x+2 \overline{)x^{2}+7x-5} \\ \end{array}$
- The next step will be looking at $\begin{array}{rl} & {\color{Red}x}+2 \overline{){\color{Red}x^{2}}+7x-5} \\ \end{array}$
- So $x \times ?? = x^{2}$
- $\begin{array}{rl} & \qquad \quad {\color{Blue}x} \\ & {\color{Blue}x+2} \overline{)x^{2}+7x-5} \\ & \qquad \;\;{\color{Red}x^{2}+2x} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x \\ & x+2 \overline{)x^{2}+7x-5} \\ & \quad \;{\color{Red}-\left(\right.}x^{2}+2x {\color{Red}\left.\right)}\\ \end{array}$
- Let's minus it
- $\begin{array}{rl} & \qquad \quad x \\ & x+2 \overline{)x^{2}+7x-5} \\ & \quad \;-\left(\underline{x^{2}+2x}\right)\\ & \qquad \qquad \quad {\color{Red}5x} \\ \end{array}$
- Bring down the next term $\begin{array}{rl} & \qquad \quad x \\ & x+2 \overline{)x^{2}+7x-5} \\ & \quad \;-\left(\underline{x^{2}+2x}\right)\\ & \qquad \qquad \quad 5x{\color{Red}-\;5} \\ \end{array}$
- Next, look at $\begin{array}{rl} & \qquad \quad x \\ & {\color{Blue}x}+2 \overline{)x^{2}+7x-5} \\ & \quad \;-\left(\underline{x^{2}+2x}\right)\\ & \qquad \qquad \quad {\color{Blue}5x}-5 \\ \end{array}$ Thus $\begin{array}{rl} & \qquad \quad x + {\color{Red}5}\\ & {\color{Blue}x}+2 \overline{)x^{2}+7x-5} \\ & \quad \;-\left(\underline{x^{2}+2x}\right)\\ & \qquad \qquad \quad {\color{Blue}5x}-5 \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x + 5\\ & x+2 \overline{)x^{2}+7x-5} \\ & \quad \;-\left(\underline{x^{2}+2x}\right)\\ & \qquad \qquad \quad 5x-5 \\ & \quad \qquad\;\;{\color{Red}-\left(\underline{5x+10}\right)}\\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x + 5\\ & x+2 \overline{)x^{2}+7x-5} \\ & \quad \;-\left(\underline{x^{2}+2x}\right)\\ & \qquad \qquad \quad 5x-5 \\ & \qquad \quad\;\;-\left(\underline{5x+10}\right)\\ & \qquad \qquad \qquad \;\;{\color{Red}-15} \\ \end{array}$
$\left(x^{3}-1 \right) \div \left( x -1 \right)$
- $\begin{array}{rl} & x-1 \overline{)x^{3}-1} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad {\color{Red}x^{2}} \\ & x-1 \overline{)x^{3}-1} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x^{2} \\ & x-1 \overline{)x^{3}-1} \\ & \quad {\color{Red}-\left( x^{3}-x^{2}\right)} \\ \end{array}$
- $\begin{array}{rl} & x-1 \overline{)x^{3}{\color{Red}+0x^{2}+0x}-1} \\ \end{array}$ or just something like $\begin{array}{rl} & x-1 \overline{)x^{3}\qquad \qquad-1} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad {\color{Red}x^{2}} \\ & x-1 \overline{)x^{3}+0x^{2}+0x-1} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x^{2} \\ & x-1 \overline{)x^{3}+0x^{2}+0x-1} \\ & \quad \;-\underline{\left(x^{3}-x^{2}\right)} \\ & \qquad \qquad \; \; {\color{Red}x^{2}\;\;+0x} \end{array}$ You can omit the $\,0x$ if you want to here
- $\begin{array}{rl} & \qquad \quad x^{2}+{\color{Red}x}\\ & x-1 \overline{)x^{3}+0x^{2}+0x-1} \\ & \quad \;-\underline{\left(x^{3}-x^{2}\right)} \\ & \qquad \qquad \; \;x^{2}\;\;+0x \\ & \qquad \quad \; {\color{Red}-\underline{\left({x^{2}\;\;-x}\right)}} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x^{2}+x\\ & x-1 \overline{)x^{3}+0x^{2}+0x-1} \\ & \quad \;-\underline{\left(x^{3}-x^{2}\right)} \\ & \qquad \qquad \; \;x^{2}\;\;+0x \\ & \qquad \quad \; -\underline{\left({x^{2}\;\;-x}\right)} \\ & \qquad \qquad \qquad \quad {\color{Red}x \;\;\; -1} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x^{2}+x\;\;+1\\ & x-1 \overline{)x^{3}+0x^{2}+0x-1} \\ & \quad \;-\underline{\left(x^{3}-x^{2}\right)} \\ & \qquad \qquad \; \;x^{2}\;\;+0x \\ & \qquad \quad \; -\underline{\left({x^{2}\;\;-x}\right)} \\ & \qquad \qquad \qquad \quad x \;\;\; -1 \\ & \qquad \qquad \qquad {\color{Red}-\underline{\left({x\;\;\;-1}\right)}} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad x^{2}+x\;\;+1\\ & x-1 \overline{)x^{3}+0x^{2}+0x-1} \\ & \quad \;-\underline{\left(x^{3}-x^{2}\right)} \\ & \qquad \qquad \; \;x^{2}\;\;+0x \\ & \qquad \quad \; -\underline{\left({x^{2}\;\;-x}\right)} \\ & \qquad \qquad \qquad \quad x \;\;\; -1 \\ & \qquad \qquad \qquad -\underline{\left({x\;\;\;-1}\right)} \\ & \qquad \qquad \qquad \qquad \quad \;\;{\color{Red}0} \\ \end{array}$

$\left(2x+3\right)\div\left(x+2\right)$
A somewhat simpler type but easy to make careless mistakes when writing the result
- $\begin{array}{rl} & x+2 \overline{)2x+3} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad {\color{Red}2} \\ & x+2 \overline{)2x+3} \\ \end{array}$ $\begin{array}{rl} & \qquad \quad 2 \\ & x+2 \overline{)2x+3} \\ & \quad \; {\color{Red}-\left(2x+4\right)} \\ \end{array}$ $\begin{array}{rl} & \qquad \quad 2 \\ & x+2 \overline{)2x+3} \\ & \quad \;\; -\underline{\left(2x+4\right)} \\ & \qquad \qquad \;{\color{Red}-1} \\ \end{array}$
- $\frac{2x+3}{x+2} =$ $2-\frac{1}{x+2}$

$\frac{x}{2x+1}$
- $x\overline{)2x+1}$ $\begin{array}{rl} & \quad 2 \\ & x\overline{)2x+1} \\ & -\left(2x+2\right) \\ \end{array}$ Eh? Wait a minute....
- $2x+1\overline{)x}$ But $2x \times ??=x$ ?
- $\begin{array}{rl} & \qquad \quad \; {\color{Red}\frac{1}{2}} \\ & 2x+1\overline{)x} \\ & \qquad {\color{Red}-\left(x+\frac{1}{2}\right)} \\ \end{array}$ $\begin{array}{rl} & \qquad \quad \; \frac{1}{2} \\ & 2x+1\overline{)x} \\ & \qquad -\underline{\left(x+\frac{1}{2}\right)} \\ & \qquad \qquad \; -\frac{1}{2} \\ \end{array}$
- $\therefore \frac{x}{2x+1}=$ $\frac{1}{2}-\frac{1}{2\left(2x+1\right)}$

$\left(x^{3}-2x^{2}+x-7\right) \div \left(x^{2}-1\right)$
- $x^{2}-1\overline{)\;x^{3}-2x^{2}+x-7}$
- $\begin{array}{rl} & \qquad \quad \; {\color{Red}x} \\ & x^{2}-1\overline{)\;x^{3}-2x^{2}+x-7} \\ & \qquad \;{\color{Red}-\left( x^{3}- \; x\right)} \\ \end{array}$ Hmmm... no good. $\begin{array}{rl} & \qquad \quad \; {\color{Red}x} \\ & x^{2}-1\overline{)\;x^{3}-2x^{2}+x-7} \\ & \qquad \;{\color{Red}-\left( x^{3} \qquad \;\; -x\right)} \\ \end{array}$ Better.
- $\begin{array}{rl} & \qquad \quad \; x \\ & x^{2}-1\overline{)\;x^{3}-2x^{2}+x-7} \\ & \qquad \;-\underline{\left( x^{3} \qquad \;\; -x\right)} \\ & \qquad \qquad\; {\color{Red}-2x^{2} +2x} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad \; x{\color{Red}\;-\;2}\\ & x^{2}-1\overline{)\;x^{3}-2x^{2}+x-7} \\ & \qquad \;-\underline{\left( x^{3} \qquad \;\; -x\right)} \\ & \qquad \qquad\; -2x^{2} +2x \\ & \qquad \qquad {\color{Red}-\left(2x^{2}\qquad +2\right)} \\ \end{array}$
- $\begin{array}{rl} & \qquad \quad \; x\;-\;2\\ & x^{2}-1\overline{)\;x^{3}-2x^{2}+x-7} \\ & \qquad \;-\underline{\left( x^{3} \qquad \;\; -x\right)} \\ & \qquad \qquad\; -2x^{2} +2x \\ & \qquad \qquad -\underline{\left(2x^{2}\qquad +2\right)} \\ & \qquad \qquad \qquad \qquad {\color{Red}2x -9} \\ \end{array}$
- $\,x^{3}-2x^{2}+x-7=$ $\left(x^{2}-1\right)\left(x-2\right)+2x-9$

$\left(2x^{3}-1\right) \div \left(x+1\right)\left(x-2\right)$
It's impossible to do it by (Unless you want to divide it twice!)
- So first we expand the divisor $\left(x+1\right)\left(x-2\right)=x^{2}-x-2$
- Answer : $\,2x^{3}-1=$ $\left(x+1\right)\left(x-2\right)\left(2x+2\right)+6x+3$
- $\begin{array}{rl} & \qquad \qquad \quad \; 2x \; + \; 2 \\ & x^{2}-x-2 \overline{)\;2x^{3}+0x^{2}+0x-1} \\ & \qquad \qquad -\underline{\left(2x^{3}-2x^{2}-4x\right)} \\ & \qquad \qquad \qquad \qquad 2x^{2}+4x-1 \\ & \qquad \qquad \qquad \quad -\underline{\left(2x^{2}-2x-4\right)} \\ & \qquad \qquad \qquad \qquad \qquad \;\; 6x\; + 3 \\ \end{array}$

Degree

When we divide polynomial of degree with polynomial of degree , we get
- Let's look at the examples we have done to give us an idea
  - $\underbrace{x^{2}+7x-5}_{degree \; 2}= {\color{Blue}\underbrace{\left(x+2\right)}_{degree \; 1}}{\color{Red}\underbrace{\left(x+5\right)}_{degree \; 1}}{\color{Purple}\underbrace{-15}_{degree \; 0}}$
  - $\underbrace{x^{3}-1}_{degree \; 3}= {\color{Blue}\underbrace{\left(x-1\right)}_{degree \; 1}} {\color{Red}\underbrace{\left(x^{2}+x+1\right)}_{degree \; 2}}$
  - $\underbrace{x^{3}-2x^{2}+x-7}_{degree \; 3}= {\color{Blue}\underbrace{\left(x^{2}-1\right)}_{degree \; 2}}{\color{Red}\underbrace{\left(x-2\right)}_{degree \; 1}} {\color{Purple}\underbrace{+2x-9}_{degree \; 1}}$
- Quotient of degree $\,m-n$
- Remainder $\,n-1$ or less

Exercise 1

Using long division, find the quotient & remainder when the polynomial is divided by the given divisor, and rewrite the polynomial

1) $\left(x^{2}+8x-3\right)\div \left(x+2\right)$

$\mbox{Quotient }=x+6, \mbox{Remainder }=-15, x^{2}+8x-3=\left(x+2\right)\left(x+6\right)-15$

2) $\left(4x^{2}-5x+2\right)\div \left(2x-1\right)$

$\mbox{Quotient }=2x-\frac{3}{2}, \mbox{Remainder }=\frac{1}{2}, 4x^{2}-5x+2=\left(2x-1\right)\left(2x-\frac{3}{2}\right)+\frac{1}{2}$

3) $\left(x-7\right)\div \left(3x+1\right)$

$\mbox{Quotient }=\frac{1}{3}, \mbox{Remainder }=-\frac{22}{3}, x-7=\frac{1}{3}\left(3x+1\right)-\frac{22}{3}$

4) $\left(x^{3}+2\right)\div \left(x+1\right)$

$\mbox{Quotient }=x^{2}-x+1, \mbox{Remainder }=1, x^{3}+2=\left(x+1\right)\left(x^{2}-x+1\right)+1$

5) $\left(x^{3}+7x^{2}-7\right)\div \left(x^{2}+3\right)$

$\mbox{Quotient }=x+7, \mbox{Remainder }=-3x-28, x^{3}+7x^{2}-7=\left(x^{2}+3\right)\left(x+7\right)-3x-28$

6) $\left(4x^{3}-x\right)\div \left(x+3\right)\left(x-1\right)$

$\mbox{Quotient }=4x-8, \mbox{Remainder }=27x-24, 4x^{3}-x=\left(x+3\right)\left(x-1\right)\left(4x-8\right)+27x-24$

7) $\left(3x^{2}-5x+1\right)\div \left(x^{2}-x+1\right)$

$\mbox{Quotient }=3, \mbox{Remainder }=-2x-2, 3x^{2}-5x+1=3\left(x^{2}-x+1\right)-2x-2$

Equality of Polynomials

Two polynomials are equal only if all coefficients are the same

For example, if $x^{2}+3x+2 \equiv Ax^{2}+Bx+C$

$\therefore$ $\, A=1, B =3, C=2$

Note:
- Equal polynomials are identity (thus the $\equiv$ sign, but using $\,=$ is still acceptable in working), it holds for all values of $\,x$
- we DO NOT, however solve for the value of $\,x$

Substituting Suitable Values

If we are given two polynomials (albeit written in different form), we can substitute ANY value of $\,x$ to help us find any unknowns. Thus

Use values which will make some of the terms to be zero
Use other (easy to count) values if needed ( $0,-1,1,\ldots$ )
Quickly check by substituting other suitable values

Example

$x+1 \equiv A\left(x-1\right)+B\left(x+2\right)$
The thing to keep in mind is we can substitute ANY value of , we choose the easiest to calculate
- To make ${\color{Red}A\left(x-1\right)}$ zero, we use $\,x=1$
- To make ${\color{Blue}B\left(x+2\right)}$ zero, we use $\,x=-2$
- $\,2=$ $\,3B$
- $\therefore B=\frac{2}{3}$
- $\,-1=-3A$
- $\therefore A=\frac{1}{3}$
We check by substituting
- $\,LHS = 1$

$3x-2 \equiv A\left(2x-1\right)+B\left(3x+2\right)$
- $\mbox{When}\;\;x=\frac{1}{2}$
- $\frac{3}{2}-2=B\left(\frac{3}{2}+2\right)$
- $-\frac{1}{2}=\frac{7}{2}B$
- $B=-\frac{1}{7}$
- $\mbox{When}\;\; x=-\frac{2}{3}$
- $\begin{align} & -2-2=A\left(-\frac{4}{3}-1\right)\\ & -4=-\frac{7}{3}A \\ & A=\frac{12}{7}\\ \end{align}$
I can't stress how important it is to check, especially here where
- we deal with fractions and negatives
- subsequent parts of the question will likely require the values found
- the method of checking is very simple and doesn't take ten seconds

$x^{2}+1 \equiv A\left(x-1\right)\left(x+2\right)+B\left(x-1\right)\left(x+3\right)+C\left(x+2\right)\left(x+3\right)$
Note that
- Thus, when we put $\,x=1$ , both $A\left(x-1\right)\left(x+2\right)$ and $B\left(x-1\right)\left(x+3\right)$ will be zero
Try this on your own first! Answer $A=\frac{5}{2}, B=-\frac{5}{3}, C=\frac{1}{6}$
$\begin{array}{lll} \mbox{When} \;\; x=1 &x=-2 & x=-3 \\ 2=12C &5=B\left(-3\right)\left(1\right) & 10 = A\left(-4\right)\left(-1\right)\\ C=\frac{1}{6} &B=-\frac{5}{3} & A = \frac{5}{2} \\ \end{array}$
Remember to check!

$x^{2}+x-1 \equiv A\left(x-1\right)^{2}+Bx\left(x-1\right)+Cx$
$\mbox{When}\;\; x=1, 1=C$
$\mbox{When}\;\; x=0, -1=A$
(you can also use , I just don't like dealing with negative)
- $\,5=A+2B+2C$
- $\,5=-1+2B+2$
- $\,B=2$
Check with (since we can't use or )
- $\,LHS =1-1-1=-1$
- $\,RHS =-1(-2)^{2}+2(-1)(-2)+1(-1)=-1$

Comparing Coefficients

When no suitable values can be substituted, we use comparison of coefficients
Always try to compare terms which has less contributing terms (highest power & constant)

Identifying contributing terms

- term in $\,x^{2}$ will come from $\left({\color{Red}x}+2\right)\left({\color{Red}2x}-1\right)$
- term in $\,x$ will come from $\left({\color{Red}x}{\color{Blue}+2}\right)\left({\color{Blue}2x}{\color{Red}-1}\right)$
- constant from $\left(x{\color{Red}+2}\right)\left(2x{\color{Red}-1}\right)$

- term in $\,x^{3}$ will come from $\left({\color{Red}x}+1\right)\left({\color{Red}x^{2}}+3x-5\right)$
- term in $\,x^{2}$ will come from $\left({\color{Red}x}{\color{Blue}+1}\right)\left({\color{Blue}x^{2}}{\color{Red}+3x}-5\right)$
- term in $\,x$ will come from $\left({\color{Red}x}{\color{Blue}+1}\right)\left(x^{2}{\color{Blue}+3x}{\color{Red}-5}\right)$
- constant from $\left(x{\color{Red}+1}\right)\left(x^{2}+3x{\color{Red}-5}\right)$
Thus, it's normally easier to look at the highest term and the constant first.

Example

$x^{3}+3x^{2}+x+3 \equiv \left(Ax+B\right)\left(x^{2}+1\right)$
First, note that there are no simple value of that we can substitute to make any term to be zero, so we will do comparison of coefficients
- For constant
  On LHS : $\,3$
  On RHS : $\left(Ax{\color{Red}+B}\right)\left(x^{2}{\color{Red}+1}\right)$
- Comparing constant: $\,3=B$

$2x^{2}+8x^{2}+11 \equiv A\left(x+B\right)^{2}+C$
If its too difficult to see the contributing terms, we can expand the . However, it is counterproductive to expand and rearrange the whole of RHS
- $\begin{align} 2x^{2}+8x^{2}+11 & \equiv A\left(x+B\right)^{2}+C \\ & \equiv A\left(x^{2}+2Bx+B^{2}\right)+C \end{align}$
- Comparing coefficients of $\,x^{2}$ : $\,2=A$
- Comparing coefficients of $\,x$ : $\,8=2AB$
  $\,B=2$
- $\begin{align} & \therefore 11= 8+C \\ &C=3 \end{align}$

Combination of both Methods

Always do substitution BEFORE comparing since substitution is easier

Example

$x^{2}+5x \equiv A\left(1+x^{2}\right)+\left(Bx+C\right)\left(1-x\right)$
We can substitute, so do it first
- $\mbox{When}\;\; x=1, 6 = 2A, \therefore A=3$
- Compare coefficient of term in $\,x^{2}$ : $\begin{align} & 1=A-B \\ & 1=3-B \\ & B=2 \end{align}$
- Compare constant : $\begin{align} & 0=A+C \\ & 0=3+C \\ & C=-3 \end{align}$
- Finished? No, check.
  $x^{2}+5x \equiv 3\left(1+x^{2}\right)+\left(2x-3\right)\left(1-x\right)$
  $\,x=0, LHS = 0, RHS = 3+(-3)(1)=0$

Exercise 2

Find the values of $A, B,C\,$

1) $1 \equiv A\left(x+1\right)+B\left(x-1\right)$

$A=\frac{1}{2}, B=-\frac{1}{2}$
- $\begin{align} & 1 \equiv A\left(x+1\right)+B\left(x-1\right)\\ & \mbox{When } x=-1, 1=-2B, \therefore B=-\frac{1}{2} \\ & \mbox{When } x=1, 1=2A, \therefore A=\frac{1}{2}\\ & \therefore A=\frac{1}{2}, B=-\frac{1}{2}\\ \end{align}$
- Check : $x=0, 1=\frac{1}{2}-\frac{1}{2}\left(-1\right)$

2) $3x+2 \equiv A\left(x+2\right)+B\left(x-3\right)$

$A=\frac{11}{5}, B=\frac{4}{5}$
- $\begin{align} & 3x+2 \equiv A\left(x+2\right)+B\left(x-3\right)\\ & \mbox{When } x=-2, -4=-5B, \therefore B=\frac{4}{5} \\ & \mbox{When } x=3, 11=5A, \therefore A=\frac{11}{5}\\ & \therefore A=\frac{11}{5}, B=\frac{4}{5}\\ \end{align}$
- Check : $x=0, 2=\frac{11}{5}\left(2\right)+\frac{4}{5}\left(-3\right)$

3) $x-1 \equiv A\left(3x+1\right)+B\left(2x-1\right)$

$A=-\frac{1}{5}, B=\frac{4}{5}$
- $\begin{align} & x-1 \equiv A\left(3x+1\right)+B\left(2x-1\right)\\ & \mbox{When } x=-\frac{1}{3}, -\frac{4}{3}=-\frac{5}{3}B, \therefore B=\frac{4}{5} \\ & \mbox{When } x=\frac{1}{2},-\frac{1}{2}=\frac{5}{2}A, \therefore A=-\frac{1}{5}\\ & \therefore A=-\frac{1}{5}, B=\frac{4}{5}\\ \end{align}$
- Check : $x=0, -1=-\frac{1}{5}\left(1\right)+\frac{4}{5}\left(-1\right)$

4) $x^{2}-2 \equiv Ax\left(x+1\right)+B\left(x+1\right)\left(x-2\right)+Cx\left(x-2\right)$

$A=\frac{1}{3}, B=1, C=-\frac{1}{3}$
- $\begin{align} & x^{2}-2 \equiv Ax\left(x+1\right)+B\left(x+1\right)\left(x-2\right)+Cx\left(x-2\right)\\ & \mbox{When } x=0, -2=B\left(1\right)\left(-2\right), \therefore B=1 \\ & \mbox{When } x=-1, -1=C\left(-1\right)\left(-3\right), \therefore C=-\frac{1}{3}\\ & \mbox{When } x=2, 2=A\left(2\right)\left(3\right), \therefore A=\frac{1}{3}\\ & \therefore A=\frac{1}{3}, B=1, C=-\frac{1}{3}\\ \end{align}$
- Check : $x=1, -1=\frac{1}{3}\left(2\right)+1\left(-2\right)-\frac{1}{3}\left(-1\right)$

5) $x^{2}-3x+5 \equiv A\left(x-3\right)\left(x+2\right)+B\left(x+2\right)\left(x-4\right)+C\left(x-3\right)\left(x-4\right)$

$A=\frac{3}{2}, B=-1, C=\frac{1}{2}$
- $\begin{align} & x^{2}-3x+5 \equiv A\left(x-3\right)\left(x+2\right)+B\left(x+2\right)\left(x-4\right)+C\left(x-3\right)\left(x-4\right)\\ & \mbox{When } x=3, 5=B\left(5\right)\left(-1\right), \therefore B=-1 \\ & \mbox{When } x=-2, 15=C\left(-5\right)\left(-6\right), \therefore C=\frac{1}{2}\\ & \mbox{When } x=4, 9=A\left(1\right)\left(6\right), \therefore A=\frac{3}{2}\\ & \therefore A=\frac{3}{2}, B=-1, C=\frac{1}{2}\\ \end{align}$
- Check : $x=0, 5=\frac{3}{2}\left(-6\right)-1\left(-8\right)+\frac{1}{2}\left(12\right)$

6) $x^{2}+3 \equiv A\left(x+1\right)^{2}+B\left(x-2\right)\left(x+1\right)+C\left(x-2\right)$

$A=\frac{7}{9}, B=\frac{2}{9}, C=-\frac{4}{3}$
- $\begin{align} & x^{2}+3 \equiv A\left(x+1\right)^{2}+B\left(x-2\right)\left(x+1\right)+C\left(x-2\right)\\ & \mbox{When } x=-1, 4=-3C, \therefore C=-\frac{4}{3} \\ & \mbox{When } x=2, 7=9A, \therefore A=\frac{7}{9}\\ & \mbox{Comparing coefficient of } x^{2} : 1 =A+B\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad 1 =\frac{7}{9}+B , \therefore B=\frac{2}{9}\\ & \therefore A=\frac{7}{9}, B=\frac{2}{9}, C=-\frac{4}{3}\\ \end{align}$
- Check : $x=0, 3=\frac{7}{9}\left(1\right)+\frac{2}{9}\left(-2\right)-\frac{4}{3}\left(-2\right)$

7) $5 \equiv Ax^{2}+Bx\left(x-4\right)+C\left(x-4\right)$

$A=\frac{5}{16}, B=-\frac{5}{16}, C=-\frac{5}{4}$
- $\begin{align} & 5 \equiv Ax^{2}+Bx\left(x-4\right)+C\left(x-4\right)\\ & \mbox{When } x=0, 5=-4C, \therefore C=-\frac{5}{4} \\ & \mbox{When } x=4, 5=16A, \therefore A=\frac{5}{16}\\ & \mbox{Comparing coefficient of } x^{2} : 0 =A+B\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad 0 =\frac{5}{16}+B , \therefore B=-\frac{5}{16}\\ & \therefore A=\frac{5}{16}, B=-\frac{5}{16}, C=-\frac{5}{4}\\ \end{align}$
- Check : $x=1, 5=\frac{5}{16}\left(1\right)-\frac{5}{16}\left(-3\right)-\frac{5}{4}\left(-3\right)$

8) $x^{2}-5x+7 \equiv A\left(x+B\right)^{2}+C$

$A=1, B=-\frac{5}{2}, C=\frac{3}{4}$
- $\begin{align} & x^{2}-5x+7 \equiv A\left(x+B\right)^{2}+C\\ & \mbox{Comparing coefficient of } x^{2} : 1 =A\\ & \mbox{Comparing constant } : 7 =AB^{2}+C\\ & \mbox{Comparing coefficient of } x : -5 =2BA\\ & \qquad \qquad \qquad \qquad \qquad \qquad -5=2B, B=-\frac{5}{2} \\ & \therefore 7=1\left(-\frac{5}{2}\right)^{2}+C, C=\frac{3}{4} \\ & \therefore A=1, B=-\frac{5}{2}, C=\frac{3}{4}\\ \end{align}$
- Check : $x=0, 7=\left(-\frac{5}{2}\right)^{2}+\frac{3}{4}$

9) $x^{2}+2 \equiv A\left(x^{2}+3\right)+\left(Bx+C\right)\left(x+1\right)$

$A=\frac{3}{4}, B=\frac{1}{4}, C=-\frac{1}{4}$
- $\begin{align} & x^{2}+2 \equiv A\left(x^{2}+3\right)+\left(Bx+C\right)\left(x+1\right)\\ & \mbox{When } x=-1, 3=4A, \therefore A=\frac{3}{4} \\ & \mbox{Comparing coefficient of } x^{2} : 1 =A+B\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad 1= \frac{3}{4}+B, \therefore B=\frac{1}{4}\\ & \mbox{Comparing constant } : 2 =3A+C\\ & \qquad \qquad \qquad \qquad \qquad 2= \frac{9}{4}+C,\therefore C=-\frac{1}{4}\\ & \therefore A=\frac{3}{4}, B=\frac{1}{4}, C=-\frac{1}{4}\\ \end{align}$
- Check : $x=0, 2=\frac{3}{4}\left(3\right)-\frac{1}{4}\left(1\right)$

10) $x \equiv A\left(x^{2}+1\right)+\left(Bx+C\right)\left(x-2\right)$

$A=\frac{2}{5}, B=-\frac{2}{5}, C=\frac{1}{5}$
- $\begin{align} & x \equiv A\left(x^{2}+1\right)+\left(Bx+C\right)\left(x-2\right)\\ & \mbox{When } x=2, 2=5A, \therefore A=\frac{2}{5} \\ & \mbox{Comparing coefficient of } x^{2} : 0 =A+B\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad 0= \frac{2}{5}+B,\therefore B=-\frac{2}{5}\\ & \mbox{Comparing constant } : 0 =A-2C\\ & \qquad \qquad \qquad \qquad \qquad 0= \frac{2}{5}-2C, \therefore C=\frac{1}{5}\\ & \therefore A=\frac{2}{5}, B=-\frac{2}{5}, C=\frac{1}{5}\\ \end{align}$
- Check : $x=0, 0=\frac{2}{5}\left(1\right)+\frac{1}{5}\left(-2\right)$

Divisible Polynomial

When there is no remainder (remainder is zero), we say that the polynomial
can be divided or is divisible by the divisor.
- Which also means we can just write $P\left(x\right)=divisor \times Q\left(x\right)$

If polynomial of degree $\,m$ is divisible with polynomial of degree $\,n$ , we get a polynomial of degree $\, m-n$
For example $\underbrace{P\left(x\right)}_{degree \; m}=\underbrace{\left(x-a\right)}_{degree\;1}Q\left(x\right)$ , $\therefore$ degree of $Q\left(x\right)$ is $\,m-1$

Factorizing Cubic Polynomial

Lets say we know that $\, x^{3}-2x^{2}-x+2$ can be divided by $\, x+1$ , and we need to find the quotient. The logical thing to do is

First, we note that the quotient will be of degree $\,2$
Thus, we know that $\,x^{3}-2x^{2}-x+2=$ $=\left(x+1\right) \times a\; quadratic\;polynomial$
However, if we write , we end up having to write the solution for three unknowns. So instead, we try to work it out in our heads directly.
- $x^{3}-2x^{2}-x+2 = \left(x+1\right)\left({\color{Red}??}\quad{\color{Red}??}\quad{\color{Red}??}\right)$
- Similarly, for the constant, $x^{3}-2x^{2}-x{\color{Red}+2} = \left(x{\color{Blue}+1}\right)\left(x^{3} \qquad {\color{Blue}??}\right)$ Thus ${\color{Blue}??}$ is $\,+2$
- So now we have $x^{3}-2x^{2}-x+2 = \left(x+1\right)\left(x^{2}\qquad\qquad+2\right)$
- Meaning $x^{3}-2x^{2}-x+2 = \left(x+1\right)\left(x^{2}\quad{\color{Red}??}x+2\right)$
- In other words, $-2x^{2}\,=$ $1 \times x^{2}$ $+ x \times ??x$
- Thus $\,??$ is $\,-3$
- $\,x^{3}-2x^{2}-x+2 =$ $\left(x+1\right)\left(x^{2}-3x+2\right)$
- Check the $\,x$ $x^{3}-2x^{2}{\color{Purple}-x}+2 =\left({\color{Red}x}{\color{Blue}+1}\right)\left(x^{2}{\color{Blue}-3x}{\color{Red}+2}\right)$ $\,2-3=-1$ Correct.

Note :

This is a faster method, however, there is nothing wrong going back to long division if somehow you just can't work it out (especially during exams)
You MUST, however, check. (since you won't get be able to see the zero remainder like in long division which will help us confirm it is correct)
Only employ this method for cubic polynomials. Don't do this for higher degrees (quartic and so on), since you will likely end up spending more time (and more likely to make mistakes) compared to long division.