Polynomials in XII Grade

In previous article we had introduced the whole scenario of grade XII^th math and now from here we will start the proper demonstration of each and every topic one by one. Today’s topic is polynomials and their solution by using factoring technique. When the term polynomial arises then its application, rational expression also comes into the account. It is expected from students of this class like you to have the knowledge of polynomials, still for making you guys remind about polynomials, let us start a fast flash back of polynomials:
Polynomials are the mathematics expressions that include variables and constants that are related by using arithmetic operators like addition, subtraction, and multiplication.
Now as the general methodology of math expression, every equation can be of two forms either linear or non-linear. You can also play linear equations worksheets to enhance yours skills, But in both cases polynomials functions are needed to be normalized for simplifying them. Simplification of polynomials every time mean that students need to calculate the values of unknown variables of equation.
Let us start with the representation scenario of any polynomial function after that we will discuss the solution process of it;
F(x)= a_nxⁿ + a_n-1 x^n-1 + .……+ a₂x² + a₁x + a₀

Every polynomial includes three terms as:leading term, constant and integer co-efficient. Here x_n is the leading term and (a_n, a_n-1, ……., a₀) are integer co-efficient.
When any polynomial is said to be as linear then it means all the derivatives of the equation are of same order and similarly when the scenario is opposite means: all the derivatives are not of the same order then it is said to be non-linear polynomial function. visit here for more on polynomials.
For example:
X² + Y² = 8 (here both are of same order so it is Linear equation)
X² + y = 7 (here x and y have different orders, so clearly it is an example of non-linear equation)

There are 4 major elementary properties of any polynomial equation, those are as follows:
1.    Sum of two or more polynomials always results as polynomials.
2.    Product of two polynomials is also a polynomial function.
3.    When two polynomials are combined together then the result is obtained by substituting variable of the first polynomial by the second one.
4.    Suppose a polynomial is a_nxⁿ+ a_n-1x^n-1 + ... + a₂x² + a₁x + a_0,
and  na_nx^n-1 + (n-1)a_n-1x^n-2 + ... + 2a₂x + a₁ is the its derivative thenif the set of co-efficient (an, an-1, ……., a0) does not contain the integer value that time Ka is to be evaluated as k times of a.
Now the basic theme to categorize any polynomial into distinct category is the presence of number of unknown variables in the function. As the name suggests Poly means “many”, so any polynomial function may have various unknown variables. Suppose any equation includes all the derivatives of only one unknown variable so that could be called as Monomial and if two variables are taking part then the formed polynomial is Binomial. Let us learn this one with examples:
-2xy + 3x - 5z ( a trinomial, because of three unknown variables)
-10xy (monomial,)
7xy + z (binomial)

That’s all about polynomial presentation, now it’s time to move on into its solution process. There are various ways of solving any polynomial equation but in this article we are going to elaborate factoring procedure.  As we all know that factoring is a normalization technique and most often used for polynomial evaluation because the only problem while solving any polynomial equation is that it includes number of derivatives of various order, which makes its presentation a bit complex.
So let us first talk about factorization, the problems of factors was studied by you and other grade XII students in their early grades but that time you guys were dealing with linear equations and here the given equation is most of the time of non-linear form.
There are mainly two situations according to which factoring also needs to be implemented in different ways, the situations are:
If the polynomial equation is ax² + bx + c
Then when constant c is positive and other when c is negative,
So let us explore these two situations and will take suitable example for better explanation:
First when the constant c is positive in that case:    polynomial only can be factorized when there are 2 factors of product (ac) that can be added into the absolute value of b.
For example:
6x² + 11x + 3
Here a= 6, b = 11 and c = 3

Now we need to know that are there two factors of (ac= 18) whose sum is 11, the answer is yes, the sum of 2 and 9 is 11 and product is 18.
So by rewriting the equation:
6x² + 11x + 3
6x² + 9x + 2x + 3
Put the common factor out
3( 3x + 1) + 2x( 1 + 3x)
(3 + 2x)(1 + 3x)
Now let us talk about the second situation where c or a  is negative,
-800x² -800x + 600
We can rewrite this equation by taking the common factor out as:
200( -4x² – 4x + 3)
Now if here the product of a and c is taken then the the result is negative.
(ac) = -XII, now the factors are 6 and 2 but 6 should be placed as negative to result the sum as -4.
200( -4x² – 6x + 2x + 3)
200( 3( 1 – 2x) + 2x ( 1-2x))
Now here common factor is (1-2x)
200(3 + 2x)(1- 2x)
This is the way to use the factoring normalization procedure to sort out the complex polynomials. While solving any polynomial related query one thing is to be remembered always that whenever there is any common factor present in all derivatives, then make sure to keep it out. This fundamental is the key statement to sort out the complex rational expressions.
Rational expressions are the fraction form that includes complex polynomials in its numerator and denominator. Now when term polynomial arises students must gets sure that he needs to normalize the expression first by using factorization as done above. Let us take an example to make you better understand this:
(x-2)/ (x+4)  +  (x+1)/(x+6) = (11x + 32)/(x² + 10x + 24)
(x-2)/ (x+4)  +  (x+1)/(x+6) = (11x + 32)/ (x+4)(x+6)
By applying cross-multiplication in the above expression:
(x+4)(x+6) [(x-2)/ (x+4)  +  (x+1)/(x+6) ]  = (11x + 32)
(x+4)(x+6) (x-2)/ (x+4)  + (x-2)/ (x+4)  (x+1)/(x+6) = (11x + 32)
Cancel out the common terms, after that remaining term is as :
(x+6) (x-2) + (x+4) (x+1) = (11x + 32)
Multiply the above terms
X² + 4x – 12 + x² + 5x +4 = 11x +32
2x² + 9x – 8 = 11x + 32
Simplify it now by using the normal arithmetic transitions
2x² -2x -40 = 0
2(x² – x – 20) = 0
Now send 2 into the denominator of RHS.
X² – x -20 =0
(x - 5) (x + 4) = 0
So it gives, x = 5 or -4
This is how any complex rational expression can be sorted out; in this article we have explained the polynomials, factorization and rational expression. All these three terms are related to each other because for solving those similar fundamentals are required to implement like Normalization.
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In upcoming posts we will discuss about Quadratic Equations in grade XII and Correlation and causation. Visit our website for information on 12th state board syllabus Tamilnadu