Factoring polynomials

When factoring out polynomials,  we find   the  polynomial  that divide out evenly from the original polynomial  .   How to factor  Polynomials:  For this we have to find all the terms that if multiplied together we get the  original polynomial.  This is continued to all the terms until this cannot be simplified any more.  If the polynomial cannot be factored any more then the polynomial is said to be completely factored.

A factor of a polynomial is any polynomial which divides evenly into   the given polynomial   For example, x + 2 is a factor of the polynomial x^2 – 4.
The factorization of a polynomial is its representation as a product its factors. For example, the factorization of x^2 – 4 is (x – 2)(x + 2).

In mathematics, factorization or factoring is the breaking apart of a polynomial into a product of other smaller polynomials. If you choose, you could then multiply these factors together, and you should get the original polynomial.
This example shows how to factoring polynomials.    Take 3x^2 – 12x + 9.   The common term in these is 3.  So take 3 out and divide each term by 3.  We get   3 (x 2-4x + 3).

Step by step explanation as how to Factor out Polynomials is given below
Factor 4a2 + 20a – 3a - 15

The first two terms have a common factor in 4a.  The last two terms have a   common factor in 3.
We need to factor those terms out.

4a ( a + 5)  -3(a+5)

Now you have a binomial.  Each term     has a factor of (a + 5).

(4a -3) (a+5) is the factored terms.

Another example is given below on factoring of polynomials
x^2 -8x + 15

start by looking at the factor pairs of 15.  We are looking for a pair of factors which add up to equal -8.  Look for the factor pairs of  +15 so that they add up to -8    The negative factor pairs of 15 are:
-15  and  -1
 -5 and  -3
Since -5 + -3 = -8 this is the pair we are looking for and we can factor the
original expression into:  (x-5)(x-3)

Addition of matrices

Addition of Matrices is an operation done on two or more matrics according to their corresponding entries. Matrix of same order can only be added. If there orders are different then addition operation cannot be applied on them.

Adding Matrices requires addition of corresponding elements of them. For example if two Matrices are added then elements in first row and first column of the two matrices will be added together to get the element in the first row and first column of the resultant matrix.

Two Matrices of order m × n (m is number of rows and n is number of columns) can be added:

Let sum of these = C = A+B then
C11 = a11 + b11, C12 = a12 + b12….,  C21 = a21 + b21, C22 = a22 + b22 and so on.

A + B = C =  ( )
Order of matrix C will also be m × n.
If two matrix are equal but opposite in sign, then there addition will be null matrix (whose all elements are zero). For example if A = ((1@2@5)) and B = ((-1@-2@-5)) then A+B = 0 as A+B = ((1+(-1)@2+(-2)@5+(-5)))= ((0@0@0))

Example 1) Add Matrices A and B if A = ((1&9@5&2))  and B = ((2&4@5&8))
Solution) Let sum of A and B be C. order of C will be same as that of A and B i.e. 2x2.
Elements of matrix C will be: C11, C 12, C 21, C22.
C11 = A11 + B11 = 1+2,
C12 = A12 + B12 = 9+4,
C21 = A21 + B21 = 5+5,
C22 = A22 + B22 = 2+8,
C = ((C11 = A11 + B11 &C12 = A12 + B12 @C21 = A21 + B21&C22 = A22 + B22))
C =   ((1 + 2&9 + 4@5 + 5 &2 + 8))
C = ((3&13@10&10))
Example 2) Perform addition of Matrices P and Q given as:
P =   ((2&0&- 3@5&3&7@3&- 7&1)), Q =  ((- 4&6&2@5&2&5@21&- 15&- 4))
Solution) let P+Q = R
R11 = P11 + Q11 = 2 +(-4) =-2,
R12 = P12 + Q12 = 0+6 = 6,
R13 = P13 + Q12 = -3+2 =-1
R21 = P21 + Q21 = 5+5 =10,
R22 = P22 + Q22 = 3+2 =5,
R23 = P23 + Q23 = 7+5 = 12
R31 = P31 + Q31 = 3+21=24
R32 = P32 + Q32 = -7+(-15) =-22
R33 = P33 + Q33 = 1+(-4)= -3
R = ((R_11&R_12&R_13@R_21&R_22&R_23@R_31&R_32&R_33 ))
R = ((-2&6&-1@10&5&12@24&-22&-3))

Gaussjordan elimination method

With the concept of matrices one can solve the system of linear equations apart from usual algebraic methods. Even in matrices topic, a system there are different methods to solve and one such method is called as Gauss Jordan elimination method named after the two mathematicians who introduced this elimination method. It is also called in short as Gauss elimination method or as Jordan elimination.
Let us first study the concept behind the Gauss elimination method. A system of equations is expressed in the matrix form as, AX = B, where A is the matrix of the coefficients, X is the matrix of the variables and B is the matrix of the constants on the right side of the system of equations. If the augmented matrix [A¦B] can be expressed in reduced row echelon form, by the method of Gauss-Jordan, that is, in the form [I¦C] (I being the identity matrix), then X = C, which gives us the solution to the system.
Now let us do Gauss-Jordan elimination step by step, with an example for clear understanding. Let a system of equations is,


-3x + 4y + z = -5
x +3y -2z = -6
2x –y + 3z = 9

   Then,A is,

-3	4	1
1	3	-2
2	-1	3

x

y

z

X is, 

-5

-6

9

B is

The following are the steps of elimination.

-3	4	1	-5
1	3	-2	-6
2	-1	3	9

[A│B]

1	-4/3	-1/3	5/3
1	3	-2	-6
2	-1	3	9

[R₁│-3]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
2	-1	3	9

[R₂│-R₁]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
0	5/3	11/3	17/3

[R₃-2R₁]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	5/3	11/3	17/3

[R₂/(13/3]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	168/39	336/39

[R₃│-(5/3)R₂]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	1	2

[R₃/(168/39)]

1	-4/3	0	7/3
0	1	-5/13	-23/13
0	0	1	2

[R₁+(R₃/3)]

1	-4/3	0	7/3
0	1	0	-1
0	0	1	2

[R₂+(5R₃/13)]

1	0	0	1
0	1	0	-1
0	0	1	2

[R₁+(4R₂/3)]

Now the finally reduced matrix is in the form [I│C], where,

1

-1

2

           C and therefore X = 

Hence, the solution to the given system of equations is, x = 1, y = -1 and z = 2. 

Thus, we had seen the used of the elimination method introduced by Gauss and Jordon.

Learning to find the determinant value of a 3x3 matrix

Consider a matrix A. If A is a square matrix then there is some value that can be assigned to A. This is called the determinant of matrix A. It is denoted by the symbol |A| or det(A). The entries of the matrices are used and an arithmetic expression is formed out of them to come to a value of the matrix. Such square matrices are most commonly used in solving system of linear equations. If the number of variables in the system is 2, then the square matrix would be a 2x2 matrix. Similarly if the number of variables in the system is n, then the square matrix would be an nXn matrix.
Determinant of a 3x3 matrix:
The determinant of a 3x3 matrix is also called a third order determinant. If we have to solve a system of three simultaneous linear equations in three variables, we shall have to deal with third order determinants or 3x3 determinant. For real numbers a1, a2, a3, b1, b2, b3, c1, c2, c3, the symbol
|a1 a2 a3|
|b1 b2 b3|
|c1 c2 c3|

Represents the determinant of 3x3 matrix.

The numbers a1, a2, a3, b1, b2, b3, c1, c2, c3 are the entries or elements of the det.  The first, the second and the third rows of the det are respectively: a1, a2, a3; b1, b2, b3 and c1, c2, c3. The first, the second and the third columns are respectively;
a1 a2 a3
b1; b2; b3;
c1 c2 c3

The value of this determinant of 3x3 matrix is defined as:

|a1 a2 a3|
|b1 b2 b3|
|c1 c2 c3|

= a1*|b2 b3| - a2 * |b1 b3| + a3 * |b1 b2|
          |c2 c3|            |c1 c3|             |c1 c2|

= a1 * (b2c3 – b3c2) – a2 * (b1c3 – b3c2) + a3 * (b1c2 – b2c1)

The above formula can be used to find the value of any 3x3 determinant. This is not the only method used to evaluate a 3x3 determinant. We can also use the Sarrus’ expansion method. It is as follows:
Suppose D
= |a b c|
   |d e f|
   |g h i|
We write five columns like this:

a b c a b
d e f d e
g h I g h
Then we multiply the terms long the red and blue lines as shown in the picture below:

The red diagonals are aei, bfg and cdh. The blue diagonals are gec, hfa and idb.
Then D = aei + bfg + cdh – gec – hfa - idb