Quadratic factorization is the process of writing a quadratic expression in the form of a product of two linear expressions. The splitting of the middle term is a technique used to factorize a quadratic expression when its leading coefficient is not equal to one. How to understand the process of Quadratic Factorization using Splitting of Middle Term?

Let’s consider the following:

To Factor the form :ax^{2} + bx + c |
Factor : 6x^{2} + 19x + 10 |

1) Find the product of 1st and last term( a x c). |
6 x 10 = 60 |

2) Find the factors of 60 in such way that addition or subtraction of that factors is the middle term (19x)(Splitting of middle term) |
15 x 4 = 60 and 15 + 4 = 19 |

3) Write the center term using the sum of the two new factors, including the proper signs. |
6x^{2} + 15x + 4x + 10 |

4) Group the terms to form pairs - the first
twoterms and the last two terms. Factor each pair by finding common factors. |
3x ( 2x + 5)+ 2(2x + 5) |

5) Factor out the shared (common)
binomial parenthesis. |
(3x + 2) ( 2x + 5) |

Example:

Find
the factors of 6x2 - 13x
+ 6

Steps to find out Quadratic equation are as follows:

1. Divide 1st factor by 1st term

so the 1st factor will be (3x-2)

1. Divide 2nd factor by 1st term

so the 2nd factor will be (3x-2)

So the factors of quadratic equation will be

6x2 -13x +6 are (2x-3) (3x-2)

Roots of the
equation are

3x – 2 = 0 ⇒ 3x = 2 so x = 2/3

2x – 3 = 0 ⇒ 2x = 3 so x = 3/2

Roots are {2/3, 3/2}

12 x^{2} +11x -15 = 0

Solution:

12x2 -15 = - 11x

12x2 -15 + 11x = 0 ( add
+ 11x )

12x2 + 11x -15 = 0

Steps to find the factors of Quadratic equation are as follows:

1. Divide the 1st factor by 1st term

so the factor will be (3x + 5)

Now divide the 2nd factor by 1st term

so the factor will be (4x – 3)

So the factors of quadratic equation

12x2 +11x -15 are (3x + 5) (4x – 3)

Find the factors of 3x2 - 2x - 1

Solution:

3 x^ 2 - 2x - 1 = 0

⇒ 3 x^2 - 3x + x- 1 = 0

⇒ 3x (x - 1) + (x - 1) = 0

⇒ (x - 1) (3x + 1) = 0

⇒ x = 1 and x = -1/3

Product of two consecutive positive integers is 240. Find the integers

Solution : Let x and x + 1 are consecutive positive integers.

x(x + 1) = 240

x2 + x = 240

x2 + x - 240 = 0

x2 + 16x - 15x - 240 = 0

x(x + 16) - 15(x -16)= 0

(x + 16) (x -15) = 0

x = -16 and x = 15

So the positive integers are 15 and 16.

Quadratic expressions are expressions with a variable of degree 2, i.e., expressions of the form ax^2 + bx + c, where a, b, and c are constants. The quadratic equation method involves rewriting the quadratic expression as a product of two linear expressions. Here, we will focus on the technique of splitting the middle term, which is used when the leading coefficient 'a' is not equal to 1. This method involves splitting the linear term 'bx' into two terms whose sum is equal to 'bx' and whose product is equal to 'a*c'. We can then factor out the common factors in the resulting expression and write the expression as a product of two linear terms.Let's take an example to understand this method in detail.

Example 1:

Factorize the quadratic expression 6x^2 + 11x + 4 Solution:

Step 1: Multiply the leading coefficient and the constant term, which gives 6 * 4 = 24

Step 2: Find two numbers whose sum is equal to the coefficient of x, which is 11, and whose product is equal to 24. The numbers are 3 and 8, since 3 + 8 = 11, and 3 * 8 = 24

Step 3: Rewrite the quadratic expression by splitting the linear term into two terms using the numbers found in step 2. We can write

6x^2 + 3x + 8x + 4

Step 4: Factor out the common factor in the first two terms, which is 3x, and the common factor in the last two terms, which is 4. This gives us 3x(2x + 1) + 4(2x + 1)

Step 5: Notice that the two terms inside the parentheses are the same. Factor out this common factor to obtain the final factorization

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

Hence, 6x^2 + 11x + 4 factors into (3x + 4)(2x + 1) Example 2:

Factorize the quadratic expression 2x^2 - 5x - 3. Solution:

Step 1: Multiply the leading coefficient and the constant term, which gives 2 * (-3) = -6

Step 2: Find two numbers whose sum is equal to the coefficient of x, which is -5, and whose product is equal to -6. The numbers are -3 and 2, since -3 + 2 = -1, and -3 * 2 = -6

Step 3: Rewrite the quadratic expression by splitting the linear term into two terms using the numbers found in step 2. We can write

2x^2 - 3x - 2x - 3

Step 4: Factor out the common factor in the first two terms, which is x, and the common factor in the last two terms, which is -1. This gives us x(2x - 3) - 1(2x - 3)

Step 5: Notice that the two terms inside the parentheses are the same. Factor out this common factor to obtain the final factorization

(x - 1)(2x - 3)

Hence, 2x^2 - 5x - 3 factors into (x - 1)(2x - 3)

In summary, the technique of splitting the middle term involves splitting the linear term into two terms whose sum is equal to the coefficient.

**Example 1:**

Factorize x^2 + 5x + 6

**Solution:** To
factorize x^2 + 5x + 6, we need to find two numbers whose sum is 5 and product
is 6. These numbers are 2 and 3. Therefore, we can split the middle term 5x as
2x + 3x. We can then factor the expression as follows:

x^2 + 5x + 6 = x^2 + 2x + 3x + 6 = (x + 2)(x + 3)

Therefore, x^2 + 5x + 6 can be factorized as (x + 2)(x + 3).

**Example 2: **

Factorize 2x^2 + 7x + 3

**Solution:** To
factorize 2x^2 + 7x + 3, we need to find two numbers whose sum is 7/2 and
product is 3/2. These numbers are 3/2 and 1/2. Therefore, we can split the
middle term 7x as 3x/2 + x/2. We can then factor the expression as follows:

2x^2 + 7x + 3 = 2x^2 + 3x/2 + x/2 + 3 = (2x + 1)(x + 3)

Therefore, 2x^2 + 7x + 3 can be factorized as (2x + 1)(x + 3)

**Example 3: **

Factorize 4x^2 - 4x - 3

**Solution:** To
factorize 4x^2 - 4x - 3, we need to find two numbers whose sum is -4/4 = -1 and
product is -3/4. These numbers are -3/2 and 1/2. Therefore, we can split the
middle term -4x as -3x - x. We can then factor the equation as follows:

4x^2 - 4x - 3 = 4x^2 - 3x - x - 3 = (4x^2 - 3x) - (x + 3) = x(4x - 3) - (x + 3) = (4x - 3)(x - 1)

Therefore, 4x^2 - 4x - 3 can be factorized as (4x - 3)(x - 1).

These examples illustrate how the technique of splitting the middle term can be used to factorize quadratic expressions into simpler forms. It is a powerful tool that can simplify complex expressions and make them easier to work with.

**Solving quadratic equations:
**One of the most common practical uses of quadratic factorization using
splitting of middle term is in solving quadratic equations. By factoring the
quadratic expression into two linear expressions, we can easily find the roots
of the quadratic equation.

**Simplifying algebraic expressions:
**The technique of quadratic factorization using splitting of middle term can be
used to simplify complex algebraic expressions. By factoring out the common
factors, we can simplify the expressions and make them easier to work with.

**Solving optimization problems:**
In optimization problems, we often need to find the maximum or minimum value of
a quadratic expression. By factoring the give expression together, we can
easily identify the critical points and determine the optimal value.

**Solving projectile motion problems:**
In physics, the technique of factoring quadratic equations
by factorization using splitting of middle term is used to solve
projectile motion problems. By factoring the quadratic equation, we can
determine the time of flight, maximum height, and range of a projectile.

**Developing computer algorithms:**
Quadratic factorization using splitting of middle term is used in computer
algorithms, such as cryptography and error-correcting codes. By factoring large
numbers, we can develop secure encryption algorithms and error-correcting
codes.

Overall, the technique of quadratic factorization using splitting of middle term is a powerful tool with practical applications in various fields. It enables us to simplify complex expressions, solve quadratic equations, and solve optimization problems, making it an essential skill for students and professionals alike.

• Quadratic Factorization using Splitting of Middle Term

• By completing the square

• Factorization using Quadratic Formula

• Solved Problems on Quadratic Equation

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