Polynomials in Grade XI

Math article on grade XI (Algebra ii – Operations on polynomials)
Friends! In previous article we had started the discussion of Algebra ii and Absolute value is the main scenario of previous article. If you get remember then in the end of previous article the discussion of System of equation was started and we will continue its remaining part in this article. Along with System of equation, today we will also explore various operations that are usually applied on polynomials.
In previous article we already had explored the Row reduction method of solving system of linear equations and now here we are going to discuss direct solving method for system of linear equation:
For solving this we will use two newer terms that are vector presentation of matrix and inverse matrix.
As explained in previous session, a system of linear equation can be represent as AX = C
Where A refers the coefficient matrix, x represents the variables derivatives and C represents the RHS of the system. But here we will use their vector form as;
AX(vector) = C(vector)
Here x is an n-dimensional vector the elements of which represent the solution of the equations and C (vector) is the constant vector of the system of equations. Now its solution can be written in the following form:
X = A^-1 C
Here A^-1, represents Inverse matrix of co-efficient of the system.
Suppose a system is as:
-x + 3y + z = 1
2x + 5y = 3
3x + y - 2z = -2
Now write this system into standard form of AX = C, as
-1  3   1     x     1
 2  5   0     y      3
 3   1  -2    z    -2
Now find the inverse of the Matrix A, as
A^{-1 =}       -10/9  7/9 -5/9
            4/9    -1/9   2/9
         -13/9  10/9  -11/9
Now multiply both sides of equation by this inverse matrix:
On the left side, A-1 cancel out the matrix A, and on the right side:
⁼               -10/9  7/9 -5/9            1
            4/9    -1/9   2/9           3
           -13/9  10/9  -11/9        -2
=  21
  -3
    39
This is the solution of the system of linear equations that can be represented as:
X         21
Y   = -3
Z          39
This is the second way by which any system of linear equation can be sort out but one thing to notice that every defined method for solving this type of queries required the implementation of matrices and its application, so students has to make their command on all the fundamentals and operation procedures of matrices for solving queries like system of linear equations.

So Now let us begin with the topic: operations on Polynomials;
“What is a Polynomial and how it is formed?, this query is not required to be answer for grade XI students, because you guys should aware of this, before coming into this standard. So here we are directly starting the operations discussion of polynomials:
First arithmetic operation to be applied on polynomial is Addition: while using this operation, students need to implement two standard properties that are Additional property and Distributive property that implies to combine like terms, like terms refer to same variables and same exponents for the variables.
distributive property worksheets in Algebra: if an expression is as ba + ca then it can be rewritten as a(b + c)
ba + ca = a(b + c)
let us take an example to explain it more:
Suppose given polynomial is
(3X² + 7x + 8) + (5x² - 8x + 2) = 3x² + 7x + 8 + 5x² - 8x + 2
Now use the distributive property to combine like terms as
(3x² + 5x²) + (7x - 8x) + (8 + 2)
8x² – x + 10
Now move towards the next operation that is Subtraction:
This one is pretty similar as Addition operation but when two terms are added then the second polynomial terms are inversed or multiplied by minus (-) sign.
We are using the same example as above:
(3X² + 7x + 8) + [- (5x² - 8x + 2)]
Here the second polynomial is need to be subtracted from first one that’s why firstly the second polynomial is multiplied by negative sign and then the simple addition will be performed.
3X² + 7x + 8 - 5x² + 8x – 2
-2x² + 15x + 6
Third general operation, which is used to solve polynomials, is multiplication. Let us find out the simplest way to execute this operation on polynomials. While evaluating these operation students need to use above principles of addition and subtraction along with distributive property and law of exponents. Let us take an example to see the practical implementation of multiplication in between two polynomials. For more information on polynomials visit here
Example: 3x(2x -5) = (3x)(2x) – 3x(5) = 6x² – 15x
Let us take one more example where number of derivatives is more than previous one:
( 2x + 1) (3x - 4) = (2x)(3x) – (2x)4 + (1)3x  -  (1)4
= 6x² - 8x + 3x – 4
= 6x² – 5x – 4

The fourth operation to be used to solve higher order polynomials is power operation. There are various standard formulas that are need to be learnt before starting the evaluation of complex polynomials. Some of these standard formulas are as following:
(x + y)⁰ = 1
 (x + y)¹ = x + y
(x + y)² = x² + 2xy + b²
(x + y)³ = x³ + 3x²y + 3xy² + y³
(x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴
(x + y)⁵ = x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y5

Let us take an example to use some of the above formulas for solving polynomials:
(x² + 2x – 3)² = (x² + 2x – 3) (x² + 2x – 3)
= (x²) (x²) + (x²)(2x) - (x²)(3) + (x²)(2x) + (2x)(x²) - (2x)(3) – 3(x²) - (2x)(3) +9
= x⁴ + 2x³ – 3x² + 4x² – 3x² + 2x³ – 6x - 3x² – 6x + 9
= x⁴ + 4x³ – 2x² - 12x + 9
This is how the polynomial multiplication sorted out. While solving this student should know that whenever the bases are same, that time exponents are added to form a single term; This basic rule is learned in earlier classes but still most of the times students get confused while tracing the solved answer of the query.
Now the final operation we are going to explore in today’s article for polynomials is Division:
This operation includes some conditions that are needed to be taking in count while dividing two polynomials, as follows:
As you guys know that polynomials have many types as Monomials, Binomials and trinomials, so this is the reason the scenario of division process varies.
1.    First condition is:  Monomial  Monomial
To sort out this law of exponents is required:
(24x⁴y²)/ (18x²y³) =  4x²/3y²
This one is simple because cancellation of terms gets easier because of Monomial.
2.    Polynomial  monomial:-
(4x⁵ – 8x³ + 12x²)/ 6x²
= 4x⁵/6x² – 8x³/6x² + 12x²/6x²
= 2x³/3 – 4x/3 + 2
3.    Polynomial ÷ polynomial;- for this purpose students need to follow long division algorithm that includes following steps:

a. first write the dividend and divisor polynomials into their standard polynomial form after that use zero coefficients for powers of the variable that are missing in the dividend and divisor.

b. then Divide first term of the divisor into the first terms of the dividend. Put its quotient above term in the dividend.

c.  in third step, Multiply quotient by all terms of the divisor and put the products under the appropriate terms of the dividend.

d.  (change signs on bottom polynomial and add) Subtract and bring down remaining terms.

e. divide the remaining term by first term until the power of the divisor is larger than the power in the dividend

Let’s take an example to execute this algorithm to solve
divide x2 + 9x +14 by (x + 7)

          
This is all about arithmetic operations on polynomials. If there is any doubt remaining in your mind related to polynomials operations or system of linear equation then you guys have the option of using Online math tutoring service where Online expert math tutors are waiting to give you immediate assistance with detailed explanation by solving related queries. These online tutors are available for 24 hours in a day so students can interact with them whenever they want.

In upcoming posts we will discuss about Types of Events in Grade XI and isosceles triangle theorem. Visit our website for information on Karnataka board syllabus

Absolute Value in Grade XI

Hey friends! Welcome to another descriptive session of Grade XI math, in previous article you guys have learned important terms about linear equations and inequalities of Algebra 1 unit of this syllabus but today in this article we are including two initial topics of Algebra ii that are related to linear inequalities. These topics are Solving Absolute Value Equations and Systems of linear equations/inequalities. Nothing much needs to tell about inequalities because that was very well explained in previous article. So directly starts with the main topics of today’s article.

Apart form starting anything we should now What is absolute value function? with Absolute value to solve equations and inequalities, let us first discuss the main definition of Absolute value. The term Absolute value is used when the definite numerical value is important rather than its sign. For example if you say that ‘x’ is 20 m far away from the original point then this causes the distance value as 20, when absolute value of ‘x’ is defined then it does not matter that in which direction it is being traveled.

In general if x = 20 or x = -20 then its absolute value is represented as │x│= 20. Absolute value inequalities are solved by using simple method that suggests to replace the inequality symbol with = sign. Once the replacement is done then solve the equation to find the critical numbers and then divide them into intervals and test to check the right one. Suppose an equation is │x - 2│< 3

So first replace it with = sign; │x - 2│=3

Now by evaluating the above equation student will have two values, x= 5 or x = -1 these two values cause two intervals, first interval includes values less than -1, second interval includes value between -1 and 5, and the third interval have values more than 5. Now test these interval values by putting them into inequality equation, it easily gets clear that only second interval value is satisfying the equation as:

When x lies in x interval then value of x can be 0, so by putting this in equation:

        │x - 2│< 3
        │0 - 2│< 3

2<3 true

So the region of x is as -1 < x < 5

 For solving less than sign inequalities, there is a short trick as:
When Absolute value inequalities is in the following form
│Algebraic Expression│< K; here K is a positive real number. Then first rewrite the equation as:
-K <│Algebraic Expression│< K
The reason behind this short-trick preference is that whenever absolute value of quantity is less than a positive real number that time the Absolute quantity is situated in between negative and positive values of the real integer line. For example if │x│<3 then as an absolute value principle the value of x must be in 3 units but it can be of both directions so we can write this one as -3 < │x│<3.
Now if the Absolute value inequality includes a greater than symbol instead of less than to solve this one also with ease, students can follow the procedure defined below;
If │Algebraic Expression│> K; where k is real integer
Then it can be rewritten as: │Algebraic Expression│< -K
Now student have two equations ahead of just one, so solve them both and the solution will comprise the union of both Absolute inequality equation’s solution.
For example: evaluate │2x + 1 │≥ 3
First rewrite the equation as │2x + 1 │≥ 3 and │2x + 1 │≤ -3
Solve both of the inequality equation as:
as │2x + 1 │≥ 3
 2x ≥ 2
x  ≥ 2/2
x ≥ 1
Second equation: │2x + 1 │≤ -3
2x ≤ -4
X ≤ -4/2
X ≤ -2
As said before, solution is the union of these both
x ≥ 1 OR X ≤ -2
This is how any Absolute Value inequality is being resolved by using the short-tricks. Now let us move to the next session of this article and that is System of Linear Equations. Fraction of this topic is introduced in previous article but here we will do the total demonstration of it. While going through system of equations one thing for sure that student should learn the fundamentals of matrix operations because the standard applications that are used to evaluate system of equation are based on matrix analysis.
Let’s start with the actual meaning of system of linear equation: A linear system is a finite collection of linear equations that includes same number of unknown variables in it. As a standard form, a linear system of m equations and n variables is represented as:
aXIx1 + a12x2 + ……. + a1nxn = b1
a21x1 + a22x2 + ……. + a2nxn = b2
….            …                      ….         .
….            …                      ….         .
….            …                      ….         .
am1x1 + am2x2 + ……+ amnxn = bm
Tuple set ( s1, s2,…..,sn), when these values are substituted in place of x variable then linear system is called Solution set of the system.  Every existing linear system can be put into one of the following category:
No solution, Unique solution or infinite solution. Linear system is said to be consistent if its solution at-least comprise one solution and if there is not a single solution existing then the system is said to be inconsistent.
Let us take an example of solving a linear system,
If the system is as:

 x1 -  x2 + x3 - x4 = 2
x1 - x2 + x3 + x4 = 0
4x1 - 4x2 + 4x3 = 4
-2x1 + 2x2 - 2x3 + x4 = -3
Relevant matrices of the above given linear system is as:
1 -1 1 -1   2
1 -1 1 1    0
4 -4 4 0    4
-2 2 -2 1  -3
Now apply the row operations to convert the matrix into more linear form
We are applying following row eliminations to convert apply the row elimination algorithm on the matrices;
1.    R2 - R1
2.    R3 - 4R1
3.    R4 + 2R1
After applying these eliminations the resultant matrices is as :
1 -1 1 -1  2
0 0 0 2   -2
0 0 0 4   -4
0 0 0 -1   1
Still the matrix is not in unit form so further row transformations are required,
1.    (1/2)R2
2.    R3 -2R2
3.    R4 + (1/2)R2

The resultant matrix is as
1 -1 1 -1  2
0 0 0 1   -1
0 0 0 0    0
0 0 0 0    0
One more transformation is applied as R1 + R2
(1) [-1] [1] 0  1
0 0 0 (1)     -1
0 0 0 0         0
0 0 0 0         0
Now the above matrix results the following equivalent linear system.
x1 = 1 + x2 - x3
x4 = -1
In above equation x2 and x3 are free variables so they can take arbitrary constants as x2= c1 and x3 = c2. Now the system is as:
x1 = 1 + c1 - c2
x2 = c1
x3 = c2
x4 = -1
Consider c1=0, c2 = 0 then obtained solution is as
x=-1
     0
     0
    -1
Now for the homogeneous linear system AX= 0
x1 - x2 +x3 - x4 = 0
x1 - x2 +x3 + x4 = 0
4x1 - 4x2 + 4x3 = 0
-2x1 +2x2 - 2x3 +x4 = 0
Now because we have its equivalent analyzed matrix, so
1 -1 1 -1    0                             (1) [-1] [1] 0  0
1  -1 1 1     0   equivalent to0 0 0 (1)     0
4 -4 4 0     0                                        0 0 0 0         0
-2 2 -2 1    0                                       0 0 0 0         0


So these above system provides two solution set for the given linear equation system, that’s why this system is said to be a consistent system.

This is how any linear system is sorted out but there are several more ways in which any system of equation can be resolved and for that you can go through the next article of grade XI Algebra ii. In next article we will continue the scenario of Algebra ii including the rest of system of equations solution procedures.

At any stage if student feels some difficulty in solving any math query then he can immediately access the online math tutoring service where expert online math tutors are available for 24 x7 hours and student can interact with them through remote connection. This online education service is the good platform to go with because due to Internet platform all the relevant data to study is easily available and you can compete your analytical skills with other students globally by attending scheduled online math tests. To have regular lesson sessions, students can open their account on these websites after which they can access the variety of tools of these websites in their study to add more ease factor, some of these tools are online math calculator, worksheets and video aids.

In upcoming posts we will discuss about Polynomials in Grade XI and translations. Visit our website for information on Maharashtra state board books