Question

Factorize by split middle term 6x^2-5x-21

Answer 1

Answer :

6x² - 5x - 21 = (2x + 3) (3x - 7)

Step-by-step explanation :

Given quadratic polynomial,

6x² - 5x - 21

⇒ It is of the form ax² + bx + c

 where

a - coefficient of x²

b - coefficient of x

c - constant term

By sum-product pattern,

>> Find the product of quadratic term [ax²] and constant term [c]

   = (6x²) × (-21)

   = -126x²

>> Find the factors of "-126x²" in pairs

   (x) (-126x)

   (2x) (-63x)

   (-2x) (63x)

   (3x) (-42x)

   (-3x) (42x)

   (6x) (-21x)

   (-6x) (21x)

   (7x) (-18x)

   (-7x) (18x)

   (9x) (-14x)

   (-9x) (14x)

>> From the above, find the pair that adds to get linear term [bx]

  9x - 14x = -5x

>> Split -5x as 9x and -14x

   6x² - 5x - 21

   6x² + 9x - 14x - 21

>> Find the common factor

   3x(2x + 3) - 7(2x + 3)

   (2x + 3) (3x - 7)

∴ 6x² - 5x - 21 = (2x + 3) (3x - 7)

Answer 2

$\huge\sf\underline{Answer}$

Step by step explanation:-

Given:-

6x²-5x-21

To find:-

Slip the term

Solution:-

6x² - 5x -21 = 0

Finding factors

-21 × 6 = -126

1 × -126

2 × - 63

3 × - 42

6 × -21

7 × -18

9 × -14

These all are factors Now,We have to find those factors should sum is -5

9 × - 14 = -126

9 - 14 = -5

So, 9x -14x

Splitting middle term:-

6x² -5x -21

6x² +9x -14x-21

3x ( 2x + 3) -7(2x + 3)

Taking common

(2x + 3) ( 3x -7)

So, 6x²-5x-21 = (2x+3)(3x-7)

Hence factorized!!!