FACTORING QUADRATIC EXPRESSIONS BY SPLITTING THE MIDDLE TERMS

Problem 1 :

2x² – 9x - 5

Solution :

Step 1 :

Product of factors = -10

Sum of factors = -9

Step 2 :

 (-5) · 2 = -10 and (-5) + 2 = -3 ≠ -9

1 · (-10) = -10 and 1 + (-10) = -9

So, the factors are -10 and 1.

Step 3 :

= 2x² + 1x – 10x – 5

By grouping,

= (2x² + 1x) + (-10x – 5)

Step 4 :

By taking the common factor, we get

= x(2x + 1) - 5(2x + 1)

= (x – 5)(2x + 1)

Problem 2 :

3x² + 5x - 2

Solution :

Step 1 :

Product of factors = -6

Sum of factors = 5

Step 2 :

 (-2) · 3 = -6 and (-2) + 3 = 1 ≠ 5

(-1) · 6 = -6 and (-1) + 6 = 5

So, the factors are -1 and 6.

Step 3 :

= 3x² - 1x + 6x – 2

By grouping,

= (3x² - 1x) + (6x – 2)

Step 4 :

By taking the common factor, we get

= x(3x – 1) + 2(3x - 1)

= (x + 2)(3x – 1)

Problem 3 :

3x² – 5x - 2

Solution :

Step 1 :

Product of factors = -6

Sum of factors = -5

Step 2 :

 (-2) · 3 = -6 and (-2) + 3 = 1 ≠ -5

1 · (-6) = -6 and 1 + (-6) = -5

So, the factors are 1 and -6.

Step 3 :

= 3x² + 1x - 6x – 2

By grouping,

= (3x² + 1x) + (-6x - 2)

Step 4 :

By taking the common factor, we get

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

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

Problem 4 :

2x² + 3x - 2

Solution :

Step 1 :

Product of factors = -4

Sum of factors = 3

Step 2 :

Finding the factors of product and sum.

(-2) · 2 = -4 and (-2) + 2 = 0 ≠ 3

(-1) · 4 = -4 and (-1) + 4 = 3

So, the factors are -1 and 4.

Step 3 :

= 2x² - 1x + 4x – 2

By grouping,

= (2x² - 1x) + (4x – 2)

Step 4 :

By taking the common factor, we get

= x(2x – 1) + 2(2x - 1)

= (x + 2)(2x – 1)

Problem 5 :

2x² + 3x - 5

Solution :

Step 1 :

Product of factors = -10

Sum of factors = 3

Step 2 :

 (-1) · 10 = -10 and (-1) + 10 = 9 ≠ 3

(-2) · 5 = -10 and (-2) + 5 = 3

So, the factors are -2 and 5.

Step 3 :

= 2x² - 2x + 5x – 5

By grouping,

= (2x² - 2x) + (5x – 5)

Step 4 :

By taking the common factor, we get

= 2x(x – 1) + 5(x - 1)

= (x - 1)(2x + 5)

Problem 6 :

5x² – 14x - 3

Solution :

Step 1 :

Product of factors = -15

Sum of factors = -14

Step 2 :

 (-3) · 5 = -15 and (-3) + 5 = 2 ≠ -14

1 · (-15) = -15 and 1 + (-15) = -14

So, the factors are 1 and -15.

Step 3 :

= 5x² + x - 15x - 3

By grouping,

= (5x² + x) + (-15x - 3)

Step 4 :

By taking the common factor, we get

= x(5x + 1) - 3(5x + 1)

 = (x - 3)(5x + 1) 

Problem 7 :

2x² – 7x + 6

Solution :

= 2x² – 7x + 6

= 2x² – 4x - 3x + 6

= 2x (x - 2) - 3(x - 2)

= (2x - 3)(x - 2)

So, the factors are (2x - 3) and (x - 2).

Problem 7 :

5x² – 14x - 3

Solution :

= 5x² – 14x - 3

= 5x² – 15x + 1x  - 3

= 5x (x - 3) + 1(X - 3)

= (5x + 1)(x - 3)

So, the factors are (5x + 1)(x - 3).

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