FACTORING WITH VARIABLE EXPONENTS

Factories :

Problem 2 :

2^{n + 2} + 2ⁿ

Solution :

2^{n + 2} + 2ⁿ

= 2ⁿ ⋅ 2² + 2ⁿ

= 2ⁿ(4 + 1)

= 5(2ⁿ)

Problem 3 :

4ⁿ + 4³ⁿ

Solution :

4ⁿ + 4³ⁿ

= 4ⁿ + 4ⁿ ⋅ 4²ⁿ

= 4ⁿ(1 + 4²ⁿ)

Problem 4 :

6^{n + 1} - 6

Solution :

6^{n + 1} - 6

= 6ⁿ ⋅ 6¹ - 6

= 6(6ⁿ - 1)

Problem 5 :

7^{n + 2} - 7

Solution :

7^{n + 2} - 7

= 7ⁿ ⋅ 7² – 7

= 7(7^{n + 1} - 1)

Problem 6 :

3^{n + 2} - 9

Solution :

3^{n + 2} - 9

= 3ⁿ ⋅ 3² – 9

= 9(3ⁿ – 1)

Problem 7 :

5(2ⁿ) + 2^{n + 2}

Solution :

5(2ⁿ) + 2^{n + 2}

= 5(2ⁿ) + 2ⁿ ⋅ 2²

= 5(2ⁿ) + 2ⁿ ⋅ 4

= 2ⁿ[5 + 4]

= 2ⁿ(9)

Problem 8 :

3^{n + 2} + 3ⁿ⁺¹ + 3ⁿ

Solution :

3^{n + 2} + 3ⁿ⁺¹ + 3ⁿ

= 3ⁿ ⋅ 3² + 3ⁿ ⋅ 3¹ + 3ⁿ

= 3ⁿ ⋅ 9 + 3ⁿ ⋅ 3 + 3ⁿ

= 13(3ⁿ)

Problem 9 :

2ⁿ⁺¹ + 3(2ⁿ) + 2^{n - 1}

Solution :

2ⁿ⁺¹ + 3(2ⁿ) + 2^{n – 1}

= 2ⁿ ⋅ 2¹ + 3(2ⁿ) + 2ⁿ ⋅ 2^-1

= 2ⁿ(2 + 3 + 1/2)

= 2ⁿ(5 + 1/2)

= 2ⁿ(11/2)

= (2^{n – 1})11

Problem 10 :

4^x + 11(2^x) + 18

Solution :

4^x + 11(2^x) + 18

= (2²)^x + 11(2^x) + 18

= (2^x)² + 11(2^x) + 18

2^x = t

= t² + 11t + 18

= (t + 2)(t + 9)

Replacing t by 2^x, we get

= (2^x + 9) (2^x + 2)

Problem 11 :

4^x – 2^x - 20

Solution :

4^x – 2^x - 20

= (2²)^x – 2^x – 20

= (2^x)² – 2^x – 20

2^x = t

= t² – t – 20

= (t - 5)(t + 4)

Replacing t by 2^x

= (2^x - 5) (2^x + 4)

Problem 12 :

9^x + 9(3^x) + 14

Solution :

9^x + 9(3^x) + 14

= (3²)^x + 9(3^x) + 14

= (3^x)² + 9(3^x) + 14

3^x = t

= t² + 9t + 14

= (t + 2)(t + 7)

Replacing t by 3^x

= (3^x + 2) (3^x + 7)

Problem 13 :

9^x + 4(3^x) - 5

Solution :

9^x + 4(3^x) - 5

= (3²)^x + 4(3^x) – 5

= (3^x)² + 4(3^x) – 5

Let 3^x = t

= t² + 4t – 5

= (t - 1)(t + 5)

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

Problem 14 :

25^x + 5^x - 2

Solution :

= 25^x + 5^x - 2

= (5²)^x + 5^x – 2

= (5^x)² + 5^x – 2

Let 5^x = t

= t² + t – 2

= (t + 2)(t - 1)

= (5^x + 2) (5^x - 1)

Problem 15 :

49^x – 7^{x + 1} + 12

Solution :

49^x – 7^{x + 1} + 12

= (7²)^x – 7^{x + 1} + 12

= (7^x)² – 7^x (7) + 12

Let t = 7^x

= t² – 7t + 12

= (t - 4) (t - 3)

= (7^x - 3) (7^x - 4)