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)

Prior Knowledge

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.

Example

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

Common Names

Function vs Equation

Polynomials is a type of function and NOT equation.

Example

Value of Polynomial

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


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

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

Example

Degree

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)

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

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

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

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

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

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

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

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

Example

Comparing Coefficients

Identifying contributing terms

Example

Combination of both Methods

Example

Exercise 2

Find the values of A, B,C\,

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

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

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

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)

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)

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

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

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

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

Divisible Polynomial

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 to do long division (there will be no remainder), but there is a faster method.

Note :

Exercise 3

Complete the following

1) x^{3}+3x^{2}+7x+5=\left(x+1\right)\left(\right. \left(x+1\right)\left(x^{2}+2x+5\right)

2) 2x^{3}-5x^{2}+5x-6=\left(x-2\right)\left(\right. \left(x-2\right)\left(2x^{2}-x+3\right)

3) 2x^{3}+5x^{2}-1=\left(2x+1\right)\left(\right. \left(2x+1\right)\left(x^{2}+2x-1\right)

4) 6x^{3}-19x^{2}+11x+6=\left(2x-3\right)\left(\right. \left(2x-3\right)\left(3x^{2}-5x-2\right)

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