Polynomial of a Square

Using the algebraic expressions, their addition, subtraction, multiplication of  the polynomial of a squares. How to factorize some algebraic expressions. You may recall the algebraic identities using the polynomial  of a square

(x + y)2 = x2 + 2xy + y2

(x – y)2 = x2 – 2xy + y2

and x2 – y2 = (x + y) (x – y)

and their use in factorization. we shall start our study with a particular type of algebraic expression, called polynomial, of square and the terminology related to it.

Polynomial of a square in one variable and their degrees.If p(x) is a polynomial in x, The power of p(x) is called the degree of the polynomial p(x). For example, 4+ 2 is a polynomial in the variable of degree 1, 2y2 – 3y + 4 is a polynomial in the variable of degree 2, 5x3 – 4x2 + x – 2 is a polynomial in the variable of degree 3 and 7u6 – `3/2` u4+u-8 is a polynomial

in the variable u of degree 6. expression like `1/(x-1)` ,`1/(x2+2x+3) ` etc. Are not a polynomial.

Addition of polynomial of a square

Addition coefficient polynomial of a square:

We add two polynomials by adding the coefficient of  like powers.

Example 1:

Find the sum of 2x– 3x+ 5x + 3 and 4x + 6x– 6x– 1.

Solution:

The associative and distributive properties of real numbers.

(2x– 3x+ 5x + 3) + (6x– 6x+ 4x – 1) = 2x+ 6x– 3x– 6x+ 5x + 4x + 3 – 1

= 2x+ 6x– (3+6)x+ (5+4)x + 2

= 2x+ 6x– 9x+ 9x + 2.

  subtraction coefficient polynomial of a square:

We subtract polynomials like addition of polynomials.

Example 2:

Subtract 2x– 3x– 1 from x+ 5x– 4x – 6.

Solution:

Using the properties of associate and distributive.

(x+ 5x– 4x – 6) – (2x– 3x– 1) = x+ 5x– 4x – 6 – 2x+ 3x+ 1

= x– 2x+ 5x+ 3x– 4x – 6 + 1

= (x– 2x3) + (5x+ 3x2) + (–4x) + (–6+1)

= –x+ 8x– 4x – 5.

multipliying polynomial of a square

To find the multiplication of two polynomials, we use the distributive properties and the law of exponents.

Example 3:

Find the product of x– 2x– 4 and 2x+ 3x – 1 .

Solution:

(x– 2x– 4) (2x+ 3x – 1)

= x(2x+ 3x – 1) + (–2x2) (2x+ 3x – 1) + (–4) (2x+ 3x – 1)

= (2x+ 3x– x3) + (–4x– 6x+ 2x2) + (–8x– 12x + 4)

= 2x+ 3x– x– 4x– 6x+ 2x– 8x– 12x + 4

= 2x+ (3x– 4x4) + (–x– 6x3) + (2x– 8x2) + (–12x) +4

= 2x– x– 7x– 6x– 12x + 4.

Solving Quadratic Equations by Factoring

An equation which has more than one terms are squared but no higher power in terms, having the syntax, ax2+bx+c =0 where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant numerical term.

Types of quadratic equation:

1. Pure quadratic equation:

The numerical coefficient cannot be zero.  If b=0 then the quadratic equation is called as a ‘pure’ quadratic equation.

2. Complete quadratic equation:

If the equation having x and x2 terms such an equation is called a ‘complete’ quadratic equation. The constant numerical term ‘c’ may or may not be zero in a complete quadratic equation.

Example, x2 + 5x + 6 = 0 and 2×2 – 5x = 0 are complete quadratic equations.

Properties of roots of solving a Quadratic Equation:

  • The sum of roots of a quadratic equation is,

x1 + x2 = -b/a ,Where a, b are coefficients of x2 and x term respectively.

  • The multiples of the  roots of a quadratic equation is, (x1)(x2) =  c/a
  • `x=(-b+-sqrt(b^2-4ac))/(2a)`

Quadratic equation may have two roots; but in certain situation only one root can give the exact solution to a problem which is logically correct.

In the quadratic equation formula, b2-4ac is discriminant,

There are three things may occur according to the discriminant.

                        b2- 4ac > 0, the equation having two different real roots.

                        b2- 4ac = 0, the quadratic equation having two equal roots.

                        b2- 4ac < 0, the equation does not have real roots. The roots are imaginary.

Steps to solving quadratic equations by factoring method:

  • Arrange in the order of powers.
  • Given expression can be in the standard form ax² + bx + c = 0.
  • Find the greatest common factor
  • Factor any complex variables.
  • Solve the terms

Example Problems:

Example 1: Solve 29x + x² + 210

Solution:

Step 1:

Given expression can be arranged in the form an powers of orders

x² + 29x + 210

Step 2:

Given expression in the standard form ax² + bx + c = 0

x² + 29x + 210 = 0

Step 3:

Find the greatest common factor for the given expression

(x + 14) (x +15) = 0

Step 4:

Solve the factors for the expression

x+ 14 = 0 and x+ 15 = 0

Solution to the given equation is x = -14 and -15.

Example 2: Solve the quadratic equations by factoring method and find its roots 3x2-4x-4=0

Solution:

Step 1: Multiply the coefficient of x2 and the constant term,

3*-4 =-12 (product term)

Step 2: Find the factors for the product term

-12 = -6 *2 (factors -6 and 2)

-6+2 = -4

Here -4 is equal to the coefficient  x

Step 3: Split the coefficient of x as -4

3x2-4x-4=0

3x2-6x+2x-4=0

Step 4: Taking the common term 3x for the first two term and 2 for the next two terms

3x(x-2) +2(x-2) =0

(3x+2) (x-2)=0.

Now set (3x+2) =0; 3x=-2; x=-2/3

(x-2)=0; x=2.

The roots are x=-2/3 and 2.

Example 3: Solve the quadratic equations by factoring method and find its roots 3x2-4x-4=0

Solution:

Step 1: Multiply the coefficient of x2 and the constant term,

3*-4 =-12 (product term)

Step 2: Find the factors for the product term

-12 = -6 *2 (factors -6 and 2)

-6+2 = -4

Here -4 is equal to the coefficient of x

Step 3: Split the coefficient of x as -4

3x2-4x-4=0

3x2-6x+2x-4=0

Step 4: Taking the common term 3x for the first two term and 2 for the next two terms

3x(x-2) +2(x-2) =0

(3x+2) (x-2)=0.

Now set (3x+2) =0; 3x=-2; x=-2/3

(x-2)=0; x=2.

The roots are x=-2/3 and 2.

Systems of Equations

Algebra is one of the branch of mathematics which  explains the relations and properties of quantity by means of letters and other symbols. In algebra, an equation is the combination of two expressions separated by an equal sign. In the equation the both sides of the equal sign has the same value. Two or more equations to be solved simultaneously the solution must satisfy all the equations in the system .for the equations have a single solution,

Example for System of Equations

Ex:1 Solve the following system of equations:

3x+2y=11————-(1)

5x-4y=11 ————-(2)

Sol:     We have to find x and y in such a way that (1) and (2) are satisfied here simultaneously. The method consists to eliminate one of the variables using both equations, multiplying them by the right constants. More precisely, if we multiply the first equation by 2 we obtain

6x+4y = 22   ———-(3)

5x-4y = 11   ———–(2)

———————

(3) + (2)    =>                   11x= 33

X=3

Now that we have x, we can find y simply replacing any of the equations (1) or (2). We choose (1), so we get 9+2y=11,

2y = 2,

y = 1.      Hence   x = 3 y = 1

Solve the following system of equations

x + y = 11————- (1)

x – 2y = -1 ————–(2)

Sol:    We have to find x and y in such a way that (1) and (2) are satisfied here simultaneously. The method consists to eliminate one of the variables using both equation, multiplying them by the right constants. More precisely, if we multiply the first equation by 2 we obtain

2x+2y =22————(3)

x – 2y =-1—————(2)

——————–

(3) +(2) =>                   3x= 21

x =7

Now that we have x, we can find y simply replacing any of the equations (1) or (2). We choose (1), so we get 7+y=11,

y=11-7,

y=4 .

Hence x=7  y=4

Ex:3 Solve the following system of equations:

x+y=12————-(1)

2x-4y= 0———–(2)

Solution:      We have to find x and y in such a way that (1) and (2) are satisfied here simultaneously. The method consists to eliminate one of the variables using both equation, multiplying them by the right constants. More precisely, if we multiply the first equation by 2  ,we obtain

4x+4y =48   ——-(3)

2x-4y=0   ———-(2)

———————

(3) +(2)    =>                   6x= 48

X=8

Now that we have x, we can find y simply replacing any of the equations (1) or (2). We choose (1), so we get x+y=12,

y = 12 – 8

y=4

Hence the values  are ,   x=8 y=4

Anti Natural Log

Natural logarithm or log means logarithm with a base e. e is a irrational constant equal to 2.718281828.

The natural log is denoted as ln(x) or log e (x). We know that e0=1. Here in this topic we are going to see about anti natural logs. Natural logarithm  is defined for all positive real numbers. The anti natural log is otherwise called as inverse of log.

When ln x = 0.693147

then, x = anti ln (0.693147)

x = 2

Properties of Natural log:

log (x) + log (y) = log (xy) (addition property)

log (x) – log (y) = log (x/y)

elog (x) = x  if x > 0

log (ex) = x

Ant natural log – Example problems:

Example 1:

Find the value of x using anti natural log.

ln x = 1.09861229

Solution:

Given,

ln x = 1.09861223

x = anti ln (1.09861223)

x = 3

Example 2:

Find the value of x using anti natural log.

ln x = 1.38629436

Solution:

Given,

ln x = 1.38629436

x = anti ln (1.38629436)

x = 4

Example 3:

Find the value of x using anti natural log.

ln x + ln 2x = 2.07944154

Solution:

Given,

ln x + ln 2x = 2.07944154

ln 2x2 = anti ln (2.07944154)

2x2 = 8

x = 2

Example 4:

Find the value of x using anti natural log.

ln x + ln 2x = 2.89037176

Solution:

Given,

ln x + ln 2x = 2.89037176

ln 2x2 = anti ln (2.89037176)

2x2 = 18

x = 3

Example 5:

Find the value of x using anti natural log.

ln 3x + ln 2x = 3.17805383

Solution:

Given,

ln 3x + ln 2x = 3.17805383

ln 6x2 = anti ln (3.17805383)

6x2 = 24

x = 2

Example 6:

Find the value of x using anti natural log.

ln 3x + ln 2x = 3.98898405

Solution:

Given,

ln 3x + ln 2x = 3.98898405

ln 6x2 = anti ln (3.98898405)

6x2 = 54

x = 3

Determinant 3×3 Matrix

A matrix is an arrangement of numbers in rows and columns.  The matrices of same order can be added or subtracted where the resultant matrix will be of same order that is if you do any operation with matrices, the result will be a matrix.  But if you do operations on a determinant it will take a matrix to a real number.  Let the given matrix A have elements of real numbers.

That is,A = `((a, b, c), (d, c, f), (g, h, i))` . Therefore the determinant of A is given by | A | = `|[a,b,c],[d,e,f],[g,h,i]|`

This can be evaluated as follows: | A | = a `|[e,f],[h,i]|` -b`|[d,f],[g,i]| ` +c`|[d,e],[g,h]|`

= a [e `xx` I – h `xx` f] – b [d `xx` I – g` xx` f] + c [d `xx` h – g `xx` e].

Therefore, | A | = a (ei – ht) – b (di – gf) + c (dh – ge).

It is very clear from the expansion |A| will have a real number.

Therefore the determinant is an association of a matrix to a real number.

Example problems on determinant of 3 x 3 matrix.

Ex 1: Find the determinant of a 3 by 3 matrix A = `([1,2,3],[4,5,6],[7,8,9])`

Solution: |A| = `|[1,2,3],[4,5,6],[7,8,9]|`

= 1` |[5,6],[8,9]|` -2 `|[4,6],[7,9]|` +3 `|[4,5],[7,8]|` .

Here according to determinant process, consider the first row.  Then in the expansion, the first term is got by taking the product of first element in the first row with the determinant formed by eleminating the values lie a along the row of 1 and the column of 1.

For the second term change the sign of the second value and take a product with determinant that is formed by elemination the elements in its row and column.  Do the same thing for the last term.

Therefore |A| = 1 [5` xx` 9 – 8 `xx ` 6] – 2 [4 `xx ` 9 – 7` xx ` 6] + 3 [4` xx ` 8 – 7` xx` 5]

= 1 [45 – 48] – 2 [36 – 42] + 3 [32 – 35]

= – 3 – 2 (- 6) + 3 (-3)

= – 3 + 12 – 9 = 12 – 1 2 = 0.

Therefore |A| = 0.

Ex 2: Find the determinant of the 3 xx 3 matrix B = `([2,4,1],[3,6,5],[8,9,7])`

Solution: |B| =  `|[2,4,1],[3,6,5],[8,9,7]|`

= 2 `|[6,5],[9,7]|` - 4 `|[3,5],[8,7]|` + 1` |[3,6],[8,9]|`

= 2 [42 – 45] – 4 [21 – 40] + 1 [27 – 48]

= 2 [- 3] – 4 [- 19] – 21.

= – 6 + 76 – 21 = 76 – 27.

= 49.

Ex 3: Find the value of the determinant `|[2,5,4],[1,4,3],[6,8,10]|`

Solution: `|[2,5,4],[1,4,3],[6,8,10]|` = `|[1,1,1],[1,4,3],[6,8,10]|` [subtracting 2nd row from the first]

= `|[1,0,0],[1,3,2],[6,2,4]|`           [subtracting 1st column from the 2nd and 3rd]

= 1 `|[3, 2],[2,4]| ` = 12 – 4 = 8.

Practice problems on determinant of 3 x 3 matrix.

Find the determinant of the following matrices.

(i) A = `([2, 5, 1], [1, 2,-6], [0, 3, 6])`

[Ans: |A| = 93].

(ii) B = `([1, 7, 1], [2, 17, 2], [4, 37, 1])`

[Ans: |B| = - 9].

Learn Equivalent Equations

An equation is a mathematical statement that asserts the equality of two expressions .Equations consist of the expressions that is to be equated to the opposite sides of an equal sign, as in   x + 3 = 5,   9 – y =7.One use of equations is in the mathematical identities, assertions that are true independent of the values of any variables contained within them. Equations that have the same solution are called equivalent equations.                                                                              (Source.Wikipedia)

Examples for equivalent equations

Example 1:

Find the equivalent factor of the given equation: x2 – x – 12

Solution:

x2 – x – 12 = x2 – 4x + 3x – 12

= x (x–4) + 3 (x–4) = (x + 3) (x – 4)

this is equivalent factors of x2 – x – 12

Example 2:

Solve (x + 3) (x + 5) = 1 – x

Solution:

(x + 3) (x + 5) = 1 – x …………………….. (1)

x2 + 8x + 15 – 1 + x = 0
x+ 9x + 14 = 0
(x + 2) (x + 7) = 0
So x = –2 ; x = –7

The solution set = {–2, –7}.

Plug in x = -2 ,  -7 in (1)

L.H.S=> (-2+3 )(-2+5)=1-(-2)

1 (3) = 3 => R.H>S

3 =3 this is called equivalent equations

Example 3:

Solve (2x + 1) (x – 2) = 0

Solution:

(2x + 1) (x – 2) = 0 …………………….. (2)

2x + 1 = 0 or x – 2 = 0 or x = –1/2, x = 2

The solution set = {-1/2, 2}.

Subject x =2 in equation (2)

(2(2) +1)(2-2)= 5(0) =0 (Hence L.H.S = R.H.S)

Examples for equivalent equations in linear equations:

Solve the equivalent equations: 2/m + 3/n – 4/p = – 20; 2/n – 4/m + 3/p = 45; 3/m – 4/n + 2/p = 5

Solution:

Let 1/m = a, 1/n = b, 1/p = c

Then 2a + 3b – 4c = –20 ……………..(1)

–4a + 2b + 3c = 45………………………….. (2)

3a – 4b + 2c = 5 ……………………………..(3)

Consider equations (1) and (2)

(1) × 3 ⇒ 6a + 9b – 12c = –60

(2) × 4 ⇒ –16a + 8b + 12c = 180 adding

–10a + 17b = 120…………………………. (4)

Consider equations (1) and (3)

(1) ⇒ 2a + 3b – 4c = –20

(3) × 2 ⇒ 6a – 8b + 4c = 10 adding

8a – 5b = –10 …………………(5)

(4) × 8 ⇒ –80a + 136b = 960

(5) × 10 ⇒ 80a – 50b = –100 adding

86b = 860 ∴ b = 10

Substituting b = 10 in equation (5) we get

8a – 5(10) = –10; 8a = –10 + 50 = 40 ∴ a = 5.

Substituting a = 5, b = 10 in equation (3) we get

3(5) – 4(10) + 2c = 5 or –25 + 2c = 5 or 2c = 5+25 = 30 ∴ c = 15.

Substitute a = 5, b = 10, c = 15, we get n = 1/5, m = 1/10, p = 1/15.

Substituting value m = 1/5 , m = 1/10 , p = 1/15 in equation

L.H.S =2/m+3/n-4/p = 2 / (1/5) +3/(1/10)-4 (1/15)

= 10 +30 – 60 = -20 =R.H.S

Rational Indices

  Indices is nothing but the main part of power operation. Indices is said to be base to the power. Let us take an example a to the power of b is ab. here a is called index and b is called power. Hence the term ab is called indices. The rational indices are either the base or power is in rational form. In this article, we see about the rational indices with example problem.

Law – Rational Indices:

Here a, m and n are the rational form.

Case 1: Product Law

am × an = am + n

Case 2: Quotient Law

am ÷ an = am + n

Case 3: Power Law

(am)n = amn

(iv) (ab)m = am . bm

(v)    `a^(m/n)` = `a^(m* (1/n))` = `root(n)(a^m)`

(vi) ao = 1

(vii) a-1 = 1/a

Example Problem – Rational Indices:

Example 1:

Find the value of (½)3 X (½)5

Solution:

Given: of (½)3 X (½)5

Expression corresponding to the product law

Formula:  am × an = am + n

a = ½, m = 3

a = ½, n = 5

(½)3 X (½)5=  ½ (3 + 5) = (½)8

= 1/(2)8 = `(1)/(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2) = ` `(1)/(256)`

Answer: (½)3 X (½)= `(1)/(256)`

Example 2:

Solve the problem`625^(3/4)` ÷ `625^(1/2)`

Solution:

Given: `625^(3/4)` ÷ `625^(1/2)`

Expression corresponding to quotient la w

`625^(3/4)` ÷ `625^(1/2)` = `625^((3/4)- (1/2))`

`625^((3 -2)/4)`  = `625^(1/4)` =`root(4)(625)`

= `root(4)(5 * 5 * 5 * 5)`

= 5

Answer:`625^(3/4)` ÷ `625^(1/2)` = 5

Example 3:

Simply the rational indices:

a) `((4^(1/2)) * 10)/(256^(1/4))`

Solution:

Step 1: Simplify the numerator of given rational indices.

41/2 * 10 = `sqrt(4)` * 5 * 2 = 2 * 5 * 2 = 22 * 5

Step 2: Simplify the denominator

`256^(1/4)`  = `root(4)(256)` = `root(4)(4 * 4 * 4 * 4)`

= 4

Step 3: Given rational indices: `((4^(1/2)) * 10)/(256^(1/4))` = `((2^(2)) * 5)/(4)`

= `((2^(2)) * 5)/(4)` `(4 * 5)/(4)`

= 5

Answer: `((4^(1/2)) * 10)/(256^(1/4))` = 5

These are the example problem in rational indices. let do the practice practice problem in rational indices.

Practice Problem – Rational Indices:

Problem 1:

What is the value of the expression `(x^(1/2))/(x^(-2))`

Answer: x3/2

Problem 2:

What is the value of expression `(4^ (3/2))/(2^-3)`

Answer: 64