FACTORING TRINOMIALS BY SPLITTING THE MIDDLE TERM

Factor each trinomial by splitting the middle term. 

Example 1 :

(p - q)² - 6(p - q) - 16

Solution :

=  (p - q)² - 6(p - q) - 16

Let t = p - q

  =  t² - 6t - 16

  =  t² - 8t + 2t - 16

  =  t(t - 8) + 2(t - 8)

  =  (t + 2)(t - 8)

  =  (p - q + 2)(p - q - 8)

The factors are (p - q + 2) and (p - q - 8).

Example 2 :

m² + 2mn - 24n²

Solution :

  =  m² + 2mn - 24n²

  =  m² + 6mn - 4mn - 24n²

  =  m(m + 6n) - 4n (m + 6n)

  =  (m - 4n)(m + 6n)

The factors are (m - 4n) and (m + 6n).

Example 3 :

 √5 a² + 2a - 3√5

Solution :

  =  √5 a² + 2a - 3√5

√5(3√5)  =  15

  =  √5 a² + 5a - 3a - 3√5

  =  √5 a(a + √5) - 3(a + √5)

  =  (√5a - 3)(a + √5)

The factors are (√5a - 3) and (a + √5).

Example 4 : 

a⁴ - 3a² + 2

Solution :

  =  a⁴ - 3a² + 2

  =  (a²)² - 3a² + 2

Let t = a²

  =  t² - 3t + 2

  =  t² - 2t - t + 2

  =  t(t - 2) - 1(t - 2)

  =  (t - 1)(t - 2)

Substitute a² for t.

  =  (a² - 1)(a² - 2)

  =  (a + 1)(a - 1)(a² - 2)

The factors are (a + 1), (a - 1) and (a² - 2).

Example 5 : 

8m³ - 2m²n - 15mn^2.

Solution :

  =  8m³ - 2m²n - 15mn²

  =  m(8m² - 2mn - 15n²)

  =  m(8m² - 12mn + 10mn - 15n²)

  =  m[4m(2m - 3n) + 5n(2m - 3n)]

  =  m(4m + 5n)(2m - 3n)

The factors are m, (4m + 5n) and (2m - 3n).

Example 6 :

(1/x²) + (1/y²) + (2/xy)

Solution :

  =  (1/x²) + (1/y²) + (2/xy)

  =  (1/x)² + (1/y)² + 2(1/x) (1/y)

  =  [(1/x) + (1/y)]²

=  (1/x + 1/y)(1/x + 1/y)

The factors are (1/x + 1/y) and (1/x + 1/y).

Example 7 :

One number is 3 more than another number. The product of the two numbers is 54. What are the numbers?

Solution :

Let x be one number, then the other number be x + 3.

The product of two numbers = 54

x(x + 3) = 54

x² + 3x = 54

x² + 3x - 54 = 0

To solve this quadratic equation, we use the factoring method.

x² + 9x - 6x - 54 = 0

x(x + 9) - 6(x + 9) = 0

(x - 6) (x + 9) = 0

Equating each factor to 0, we get

x - 6 = 0 and x + 9 = 0

x = 6 and x = -9

Here one number x = 6, then the other number x + 3 = 9

so, the required numbers are 6 and 9.

Example 8 :

If the product of two consecutive even numbers is 168, what are the numbers ?

Solution :

Let x be the even number, its consecutive number will be x + 2.

The product of two consecutive even numbers = 168

x(x + 2) = 168

x² + 2x = 168

x² + 2x - 168 = 0

x² + 14x - 12x - 168 = 0

x(x + 14) - 12(x + 14) = 0

(x - 12)(x + 14) = 0

x = 12 and x = -14

x + 2 ==> 12 + 2 ==> 14

So, the consecutive even numbers are 12 and 14.

Example 9 :

One number is four more than another number. If the square of the smaller number is 2 less than 3 times the larger number, what are the numbers?

Solution :

Let x be another number.

One number = x + 4

x be the smaller number and x + 4 will be the larger number.

x² = 3(x + 4) - 2

x² = 3x + 12 - 2

x² = 3x + 10

x² - 3x - 10 = 0

x² - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

Equating each factor to 0, we get 

x + 2 = 0 and x - 5 = 0

x = -2 and x = 5

another number x = 5

one number = 5 + 4 ==> 9

So, the required numbers are 5 and 9.

Example 10 :

Five less than the square of a number is 44. What is the number?

Solution :

Let x be the required number.

x² - 5 = 44

x² - 5 - 44 = 0

x² - 49 = 0

x² - 7² = 0

(x + 7) (x - 7) = 0

x = -7 and x = 7

So, the required number is -7 and 7.

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