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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 :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 6x^{2} - 13x + 66x ^{2} - 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

1) 12x

12x

12x

12x

12x

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

3x

⇒ 3x

⇒ 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.

Let x and x + 1 are consecutive positive integers.

x(x + 1) = 240

x

x

x

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 Factorization using Splitting of Middle Term

• By completing the square

• Factorization using Quadratic Formula

• Solved Problems on Quadratic Equation

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