Factoring Quadratic Trinomials

Introduction

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials. In mathematics, factorization (also factorisation in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. In this article we shall discuss about factoring quadratic trinomial.

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Example Problem on Factoring Trinomials

Problem 1:

Factoring trinomials

15 – 2x – x2.

Solution: Writing in the standard form,

15 – 2x – x2 = –x2 – 2x + 15

= (–1) (x2 + 2x – 15).

Here, we find –15 = 5 × –3, 5 + (–3) = 2

Hence, we get 15 – 2x – x2 = (–1) [(x+5) {x + (–3)}]

= (–1) (x +5)(x – 3)

= (x + 5) ( 3 – x).

Problem 2:

Factoring  trinomials

x2 – x – 132.

Solution: We find –132 = (–12) × (11), (–12) + 11 = –1.

Hence we get x2 – x – 132 = [x + (–12)] (x + 11) = (x – 12) (x + 11).

Next, we consider the quadratic polynomial ax2 + bx + c where a, b, c are integers and a ≠0,1. If we are able to find two integers p and q such that pq = ac and p + q = b. Then

ax2 + bx + c =`1/a ` (a2x2 + abx + ac)

=`1/a` [a2x2 + a(p+q)x + pq]= `1/a` [a2x2 + apx + aqx + pq]=`1/a` [ax (ax + p) + q(ax + p)]

=`1/a` (ax + p) (ax + q)

Thus, we are able to factorize the expression

Problem 3:

Factoring trinomial

2x2+ 7x + 3.

Solution: Here a = coefficient of x2 = 2

b = coefficient of x = 7

c = constant term = 3

We find a × c = 2 × 3 = 6 = 6 × 1, 6 + 1 = 7 = b. Hence

2x2 + 7x + 3 = 21 (2x + 6) (2x + 1) =(x+3)(2x+1).

Instead of applying the final result of the rule, we can also do the factorization by splitting the middle term and grouping as follows:

2x2 + 7x + 3 = 2x2 + (6 + 1)x + 3

= 2x2 + 6x + x + 3

= 2x(x + 3) + (1)(x+3) = (2x+1) (x+3).

Practice Problem on Factoring Quadratic Trinomials

Problem 1:

Factorize 8a2 + 2a – 3.

Answer:

(4a + 3) (2a – 1)

Problem 2:

Factorize 6 + 11/2 x + x2.

Answer:

1/2 (x + 4) (2x + 3).

Multiplying and Dividing Polynomials

Multiplying  and dividing polynomials:

            The polynomial is one or more terms having the constants with a single variable or a combination of a variable with the integral powers.These terms are with addition or subtraction. Multiplying polynomial is a method in which each term of the polynomial are multiplied to each other one by one. To multiply we can use either horizontal method or vertical method. The polynomial division is done by long division method. Dividing polynomials can be factorization method. For that we have to factorize the given polynomial. Let's see the steps for multiplying and dividing polynomials and examples.

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Steps for Multiplying Polynomials:

Step 1: First write the factors inside the parenthesis.

Step 2: If one of the polynomial is shorter than the other then write the small polynomial as a first one.

Step 3: Now take the first term from first polynomial and multiply it with the each term of the second polynomial.

Step 4: Now take the next term from the first polynomial and multiply it with the each  term of the second polynomial.

Step 5: Repeat the step 4 until the end of the last term.

Step 6: Now add all the solution which we got in previous steps.

Step 7: Combine the like terms.

Step 8: Simplify the solutions by combining common terms.

Example for multiplying polynomials:

Let us consider the given polynomial are (x^2 +1) and (x+1)

The given polynomial can be written as (x+1) (x^2 +1)`=>` x*x^2+x*1+1*x^2+1*1

`=>` x^3+x+x^2+1

The above expression can be written as x^3+x^2+x+1

Steps for Dividing Polynomials:

Step 1:Set up the given polynomials into long division.

Step 2:To get the first term of the quotient divide the first term of the divisor by the first term of the dividend.

Step 3: Multiply the divisor with the term we got from the step 2.

Step 4:Subtract the divisor from the result we got from step 3.

Step 5: We have to do the steps 2, 3 and 4 until there is no term to divide.

Step 6:The answer is the quotient and if we have remainder write the remainder over the Quotient.

Example for dividing polynomials:

Let us consider the given polynomial (x^2-x-2) and (x+1)

In this the quotient is x-2 and the remainder is 0.

Sets and Relations

Introduction about Sets and Relations:

           A group of elements is called a set when the elements in the group are distinct. A characteristic of two objects is called Relation.

Example for Set:

            The collection of all natural numbers
            The collection of all equilateral triangles in a plane.
   Let us learn more about sets and relations in this chapter.

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More about Sets:

           An object of a set is called an element or a member or an individual of the set. We usually denote a set by an upper-case letter like A or B and an element of a set by a lower-case letter such as x or y. If x is an element of a set A, we indicate this fact symbolically by writing    x ? A. The symbol ? stands for ‘is an element of’ or ‘is a member of ’or ‘belongs to’. When an object x is not an element of the set A, we denote this fact by writing x ? A. Here, the symbol ? stands for ‘is not an element of’ or ‘is not a member of’ or ‘does not belong to’.

             For example, if A is the set {1, 3, 4, 5}, then the elements of A are 1, 3, 4 and 5; that is, we have 1 ? A, 3 ? A, 4 ? A and 5 ? A. We note that 6 ? A, -11 ? A, 9 ? A,

Types of Relations:

A relation is said to be a reflexive for the each element ‘a’ ? A. (a R a) Here R represents the relation.
A relation is said to be a reflexive for the each element ‘a’?  A. The element a is not related to a.
A relation is said to be a symmetric for the element a ~ b then b ~ a.
A relation is said to be a anti symmetric for the element a~ b then b  a.
A relation is said to be a transitive for the element aRb and bRc then aRc

Introduction of monomial functions

Introduction of monomial functions :

A term is a variable or a number (which is constant) or a variable with a number attached (called a co-efficient). A single tem is called a monomial. Monomial can be written as f(x) = C (or) f(x) = C. x ⁿ

Where n is a positive integer

k is a constant

How to Solve Monomial Functions:

• To add monomial

You can only add monomial that has the same base. If the bases are same, then we can add their co-efficient.

• To subtract monomial

Like in addition, for subtraction also the monomial should be of same bases. The co-efficient is subtracted then.

• To multiply monomial

If there are coefficients on the terms then we must first multiply coefficients by coefficients. Then we can multiply the variable of one term by the other.

• To divide monomial

When we divide the monomial, remember that the numerical coefficients are divided and then the literal coefficients (such as a and b).

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Sample Problems for Monomial Functions:

Pro 1:  Add the following monomial 5x, 2y, 3x

Sol :    Here we have two different terms x and y. We can add only the coefficients of same monomial.

            = (5x + 3x) + 2y

            = 8x + 2y

Pro 2 :  Add the monomial 3x³+ 2y³, 3xy, 2xy, 3y³

Sol :     3x³ + 2y³ + 3xy + 2xy + 3y³

            3x³ + 2y³ + 3xy + 2xy + 3y³

            3x³ + 5y³ + 5xy

Pro 3:   Subtract the following: 2x from 5x

Sol :    The bases are same. Hence we can subtract

            = 5x – 2x

            = 3x

Pro 4:   Solve monomial 3x³ + 2y³- 3xy-2xy- 3y³

Sol :      = 3x³ + 2y³ - 3xy - 2xy - 3y³

             = 3x³ – 5xy - y³

Pro 5 :  Multiply the following monomial: (3x)(4y)

Sol :    First multiply the coefficients

    (3)(4) = 12

   Then multiply the variables

   (x)(y) = xy

  So the answer is: 12xy

Pro 6 :  Multiply the following monomial: (8x³)(2x⁵)

Sol :    First multiply the coefficients

      (8)(2) = 16

    Multiplying the variables

      (x³)(x⁵) = (x³⁺⁵) = x⁸

   So the answer is 16x

Pro 7 :   Divide 4ax by 2ay

Sol :    First: 4 / 2 = 2

  Second: ax / ay = x/y

   The solution is 2x/y

Pro 8 :  Divide (2a² bc) by  (4ac)

Sol  :    First: 2/4 = 1/2

      Second: a² bc / ac = ab

     The solution is (1/2) ab = ab / 2

Introduction to Binomial Model Statistics

Introduction to Binomial Model Statistics

           A binomial random variable be the number of success x in n frequent trials of a binomial experiment. The probability distribution of a binomial random variable is calling a binomial distribution. This is also recognized as a Bernoulli distribution).

` sum_(x=0)^nP(x=x)=sum_(x=0)^n(nC_(x)p^(x)q^(n-x)) ` . 

Where x- number of successes that answer as of the binomial testing.

             n- Number of trials during the binomial testing.

             P- Probability of success on an individual trial.

             q-   Probability of failure on an individual trial. (equal to 1 - P.)

Properties for Binomial Model Statistics

The testing consists of n repetitive trials.
Every one trial is able to result in just two possible outcomes. We describe one of these outcomes a success in addition to the other, a failure.
The probability of success, denote by P, is the identical on every trial.
The trials be independent so as to is, the result on one trial do not change the result on other trials.
Problems for Binomial Model Statistics

Problem 1 for Binomial Model Statistics

        6 cards are drawn successively with alternate from well shuffle deck of 52 cards. What is the probability that

               i)  all the six cards are spades

               ii) only 3 cards are spades

              iii) none is a spade.

Introduction for angle sum property of quadrilateral

Introduction for angle sum property of quadrilateral:

              A quadrilateral is a two-dimensional figure created by connecting four segments endpoint to endpoint with each segment intersecting exactly two others. It also has four sides and four angles. Sum of interior angle of any polygon is (n-2) 180^o. Here quadrilateral has four sides so interior angle is (4-2)180^o = 360^o.Exterior angle of any polygon is 360^o so the exterior angle of a quadrilateral is 360^o.

Types of Quadrilaterals

• Trapezoid
• Parallelogram
• Square
• Rectangle
• Rhombus 

 

Angle Sum Property of Quadrilateral- Trapezoid

• The sum of the adjacent angles are equal to 180^o
• The sum of all the interior angles are equal to 360^o 

• In the above figure  the sum of two adjacent angles are equal to 180^o
        100 + 80 = 180
        100 + 80 = 180
• The sum of all the internal angles are equal to 360^o
         100 + 100 + 80 + 80 = 360

        

Angle sum property of Parallelogram:

• The opposite angles of a parallelogram are equal
• The opposite sides of a parallelogram are parallel

 

• In the above diagram The sum of all the interior angles are equal to 360^o

               110 + 70 + 110 + 70 = 360

• The sum of the adjacent angles are equal to 180^o
               110 + 70 = 180
            110 + 70 = 180

Angle sum property of Square

• All the four sides of a square are equal
• All the four angles of a square are equal, And the angles are equal to 90^o

• The sum of all the internal angles are equal to 360^o

     

• In the above diagram, All the four angles of a square are equal, And the angles are equal to 90^o
                    90 + 90 + 90 + 90 = 36^o

Angle Sum Property of Quadrilateral- Rectangle

• All the four angles of a square are equal, And the angles are equal to 90^o

• The sum of all the internal angles are equal to 360^o

    

• In the above diagram, The sum of all the internal angles are equal to 360^o
               90 + 90 + 90 + 90 = 360^o

Angle sum property of Rhombus

• The sum of all the internal angles are equal to 360^o

• Opposite angles of a rhombus are equal 

• In the above diagram, The sum of all the internal angles are equal to 360^o
                     115 + 65 + 115 + 65 = 360