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Basics for polynomials

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  a value. So in P(0) , all  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 :)

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