# FACTORING SPECIAL CASES

Factoring can be done in the following ways.

Here we see, how we are factoring using algebraic identities.

a2 + 2ab + b2 = (a + b)2 ==> (a + b) (a + b)

a2 - 2ab + b2 = (a - b)2 ==> (a - b) (a - b)

a2 - b2 ==> (a + b) (a - b)

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

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