FACTORING SPECIAL CASES

Factor each completely.

Problem 1 :

16n² - 9

Solution:

The polynomial is a difference of two squares.

= (4n)² - 3²

= (4n + 3) (4n - 3)

Problem 2 :

4m² - 25

Solution:

The polynomial is a difference of two squares.

= (2m)² - 5²

= (2m + 5) (2m -5)

Problem 3 :

16b² - 40b + 25

Solution:

Since the first terms and last terms are perfect squares. The trinomial is a perfect square. 

= 16b² - 40b + 25

a = 4b, b = 5

Write the trinomial as a² - 2ab + b²

= (4b)² - 2(4b)(5) + 5²

= (4b - 5)²

So, the linear factors are (4b - 5)(4b - 5).

Problem 4 :

4x² - 4x + 1

Solution:

The trinomial is a perfect square. Factor

= 4x² - 4x + 1

a = 2x, b = 1

Write the trinomial as a² - 2ab + b²

= (2x)² - 2(2x)(1) + 1²

= (2x - 1)²

So, the linear factors are (2x - 1)(2x - 1).

Problem 5 :

9x² - 1

Solution:

The polynomial is a difference of two squares.

= (3x)² - 1²

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

Problem 6 :

n² - 25

Solution:

The polynomial is a difference of two squares.

= n² - 5²

= (n + 5) (n - 5)

Problem 7 :

n⁴ - 100

Solution:

The polynomial is a difference of two squares.

= (n²)² - (10)²

= (n² + 10) (n² - 10)

Problem 8 :

a⁴ - 9

Solution:

The polynomial is a difference of two squares.

= (a²)² - (3)²

= (a² + 3) (a² - 3)

Problem 9 :

k⁴ - 36

Solution:

The polynomial is a difference of two squares.

= (k²)² - (6)²

= (k² + 6) (k² - 6)

Problem 10 :

n⁴ - 49

Solution:

The polynomial is a difference of two squares.

= (n²)² - (7)²

= (n² + 7) (n² - 7)

Problem 11 :

98n² - 200

Solution:

The polynomial is a difference of two squares.

= 98n² - 200

= 2(49n² - 100)

= 2((7n)² - (10)²)

= 2(7n + 10) (7n - 10)

Problem 12 :

3 + 6b + 3b²

Solution:

= 3 + 6b + 3b²

= 3(1 + 2b + b²)

= 3(1 + b)²

So, the linear factors are 3 (1 + b)(1 + b).

Problem 13 :

400 - 36v²

Solution:

= 400 - 36v²

= 4(100 - 9v²)

= 4(10)² - (3v)²

= 4(10 + 3v) (10 - 3v)

Problem 14 :

100x² + 180x + 81

Solution:

The trinomial is a perfect square. Factor

= 100x² + 180x + 81

a = 10x, b = 9

Write the trinomial as a² + 2ab + b²

= (10x)² + 2(10x)(9) + 9²

= (10x + 9)²

So, the linear factors are (10x + 9)(10x + 9).

Problem 15 :

10n² + 100n + 250

Solution:

The trinomial is a perfect square. Factor

= 10n² + 100n + 250

= 10(n² + 10n + 25)

a = n, b = 5

Write the trinomial as a² + 2ab + b2

= 10((n)² + 2(n)(5) + 5²)

= 10(n + 5)²

So, the linear factors are 10(n + 5)(n + 5).

Problem 16 :

49n² - 56n + 16

Solution:

The trinomial is a perfect square. Factor

= 49n² - 56n + 16

a = 7n, b = 4

Write the trinomial as a2 - 2ab + b2

= (7n)² - 2(7n)(4) + 4²

= (7n + 4) (7n - 4)

Problem 17 :

49x² - 100

Solution:

The polynomial is a difference of two squares.

= (7x)² - (10)²

= (7x + 10) (7x - 10)

Problem 18 :

1 - r²

Solution:

The polynomial is a difference of two squares.

= 1² - r²

= (1 + r) (1 - r)

Problem 19 :

10p³ - 1960p

Solution:

= 10p³ - 1960p

= 10p(p2 - 196)

= 10p((p)² - (14)²)

= 10p(p + 14) (p - 14)

Problem 20 :

343b² - 7b⁴

Solution:

= 343b² - 7b⁴

= 7b² (7² - b²)

= 7b²(7 + b) (7 - b)

Problem 21 :

81v⁴ - 900v²

Solution:

= 81v⁴ - 900v²

= 9v²(3v² - 10²)

= 9v²(3v + 10) (3v - 10)

Problem 22 :

200m⁴ + 80m³ + 8m²

Solution:

= 200m⁴ + 80m³ + 8m²

= 8m²(25m² + 10m + 1)

a = 5m, b = 1

Write the trinomial as a² + 2ab + b2

= 8m²(5m)² + 2(5m)(1) + 1²

= 8m²(5m + 1)²