Quadratic Factorization using Splitting of Middle Term

Quadratic factorization using splitting of middle term : In this method splitting of middle term in to two factors.



In Quadratic Factorization using Splitting of Middle Term which is x term is the sum of two factors and product equal to last term.
To Factor the form :ax2 + bx + c Factor : 6x2 + 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.
6x2 + 15x + 4x + 10
4) Group the terms to form pairs - the first two
terms 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)


Quadratic Factorization using Splitting of Middle Term

Example: Find the factors of 6x2 - 13x + 6
6x2 - 13 x + 6 ----->(1)
a.c = Product of 6 and 6 = 36
Factors of 36 = 2,18
= 3,12
= 4,9
Only the factors 4 and 9 gives 13-->(4 + 9)
For -13 , both the factors have negative sign. – 4 – 9 = - 13
Equation (1) ⇒ 6x 2 - 4x – 9x + 6
⇒2x ( 3x – 2 ) – 3 ( 3x – 2 )
(3x – 2 ) ( 2x – 3) are the factors.



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}

Examples on Quadratic Factorization using Splitting of Middle Term

1) 12x
2 -15 = 11x

Solution :
12x
2 -15 = - 11x

12x
2 -15 + 11x = 0 [ add + 11x

12x
2 + 11x -15 = 0

12x
2 + 20x - 9x -15 = 0

4x(3x + 5) - 3(3x + 5) = 0

(3x + 5)(4x - 3) = 0

3x + 5 = 0 or 4x - 3 = 0

3x = - 5 or 4x = 3

x = -5/3 or x = 3/4

Solution is (-5/3,3/4)

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2) Find the factors of 3x
2 - 2x - 1

Solution :
3x
2 - 2x - 1 = 0

⇒ 3x
2 - 3x + x- 1 = 0

⇒ 3x(x - 1) + (x - 1) = 0

⇒ (x - 1)(3x + 1) = 0

⇒ x = 1 and x = -1/3

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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

x
2 + x = 240

x
2 + x - 240 = 0

x
2 + 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.

Introduction of Quadratic Equations

Quadratic Factorization using Splitting of Middle Term
By completing the square
Factorization using Quadratic Formula
Solved Problems on Quadratic Equation

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