 ask-math.com 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.