Basic Formulas on Polynomials

Below are several formulas that are useful for Sec 2 students, those marked with a * is additional information and are not frequently seen in Sec 2 syllabus.

Formula for solving roots for quadratic equations

For quadratic equations in the form $ax^2+bx+c=0 (a,b,c \in R, a \not= 0)$ , it has solutions

$x_{1,2} = \displaystyle{\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$

The discriminant of the quadratic equation, $D=b^2-4ac$ , determines the properties of the roots.

For $D<0$ , the solutions are complex, in other words, the quadratic equation has no real roots.

For $D=0$ , the solutions degenerate to one real solution, in other words, the quadratic equation has two equal real roots.

For $D>0$ , the solutions are distinct and real, in other words, the quadratic equation has two distinct real roots.

The Binomial Theorem

For $n \in N$ , we can derive the formula:

$(x+y)^n=\displaystyle{\sum_{i=0}^n \binom{n}{i}x^{n-i}y^i}$

where $\displaystyle{\binom{n}{i}}$ is the binomial coefficient, satisfying the conditions of $n,i \in N_0, i \leq n$ , and is defined as

$\displaystyle{\binom{n}{i}} = \displaystyle{\frac{n!}{(n-i)!i!}}$ .

Also, an interesting fact to note is that

$\displaystyle{\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n}$

This implies that the sum of binomial coefficients of the expression $(x+y)^n$ is $2^n$ . It also comes with an interesting proof:

Let $x,y$ in the binomial formula be 1, which leads us to

$\displaystyle{(1+1)^n=\sum_{i=0}^n\binom{n}{i}1^{n-i}1^i}$

Since the condition for the binomial formula to be true is $n,i \in N_0$ and $i \leq n$ , we can simplify the expression to:

$\displaystyle{2^n=\sum_{i=0}^n\binom{n}{i}}$ , proven.

The Factor Theorem

A polynomial $P(x)$ is divisible by the binomal $(x-a)$ if and only if $P(a) = 0$ . This is quite self-explanatory, so I won’t elaborate on this. Ok… I change my mind. $P(x)$ is a function of P. Usually, $F(x)$ is used.

When you put something in a function, what it really means is you are giving $x$ a value. So in $P(0)$ , all $x$ become 0 and whatever is left of the function is the remainder. I will explain more under remainder theorem. In $P(a) = 0$ , there is no remainder and therefore (x-a) is a multiple of the polynomial.

The Remainder Theorem

It is also worth noting that $P(x)=(x-a)q(x)+r(x)$ , where $P(x)$ is a polynomial, $(x-a)$ is a linear factor, where $a$ is just some number. Also, $q(x)$ is the quotient polynomial when $P(x)$ is divided by the factor $(x-a)$ , while $r(x)$ is the remainder.

That is all under the topic of polynomials, other formulas will be written in detail in the respective future posts.

Cheers,

etzhky

Basics for polynomials