FACTORISE INVOLVING EXPONENTIAL FUNCTIONS

Factorise the following.

Example 1 :

3ⁿ⁺² - 9

Solution :

Given, 3ⁿ⁺² – 9

By using the product rule of exponents

a^m+n  =  a^m . aⁿ 

=  [(3ⁿ . 3²) – 9]

=  [(3ⁿ . 9) – 9]

=  9(3ⁿ – 1)

So, factored form is 9(3ⁿ – 1)

Example 2 :

5(2ⁿ) + 2ⁿ⁺²

Solution :

=  5(2ⁿ) + (2ⁿ . 2²)

=  5(2ⁿ) + (2ⁿ . 4)

=  5(2ⁿ) + 4(2ⁿ)

=  9(2ⁿ)

So, factored form is 9(2ⁿ)

Example 3 :

3ⁿ⁺² + 3ⁿ⁺¹ + 3ⁿ

Solution :

=  (3ⁿ . 3²) + (3ⁿ . 3¹) + 3ⁿ

=  9(3ⁿ) + 3(3ⁿ) + 1(3ⁿ)

=  13(3ⁿ)

So, factored form is 13(3ⁿ)

Example 4 :

2ⁿ⁺¹ + 3(2ⁿ) + 2^n-1

Solution :

2ⁿ⁺¹ + 3(2ⁿ) + 2^n-1

Using the rule a^m-n  =  a^m/aⁿ

=  (2ⁿ . 2¹) + 3(2ⁿ) + (2ⁿ/2¹)

=  2(2ⁿ) + 3(2ⁿ) + (2ⁿ/2¹)

=  2ⁿ(2+3+1/2)

=  2ⁿ[(4+6+1)/2]

=  2ⁿ[11/2]

=  11(2^n-1)

Example 5 :

4^x + 8(2^x) + 15

Solution :

=  4^x + 8(2^x) + 15

By using the power rule in exponents (a^x)^y  =  a^xy, we get

=  (2²)^x + 8(2^x) + 15

=  (2^x)² + 8(2^x) + 15

Let u  =  2^x,

=  u² + 8u + 15

Now, we have the quadratic equation form,

u² + 8u + 15

By factorization, we get

(u + 5) (u + 3)

By replacing u  =  2^x, we get

(2^x + 5) (2^x + 3)

So, the factored form is (2^x + 5) (2^x + 3)

Example 6 :

4^x + 6(2^x) - 7

Solution :

=  (2²)^x + 6(2^x) - 7

=  (2^x)² + 6(2^x) - 7

Let u  =  2^x,

=  u² + 6u - 7

Now, we have the quadratic equation form,

u² + 6u - 7

By factorization, we get

(u + 7) (u - 1)

Replacing 2^x, we get

(2^x + 7) (2^x - 1)

So, factored form is (2^x + 7) (2^x - 1)

Example 7 :

9^x - 3(3^x) - 10

Solution :

Given, 9^x - 3(3^x) - 10

=  (3²)^x - 3(3^x) - 10

=  (3^x)² - 3(3^x) - 10

Let u  =  3^x,

=  u² - 3u - 10

Now, we have the quadratic equation form,

u² - 3u - 10

By factorization, we get

(u + 2) (u - 5)

By replacing u  =  3^x, we get

(3^x + 2) (3^x - 5)

So, factored form is (3^x + 2) (3^x - 5)

Example 8 :

9^x - 6(3^x) + 8

Solution :

Given, 9^x - 6(3^x) + 8

=  (3²)^x - 6(3^x) + 8

=  (3^x)² - 6(3^x) + 8

Let u  =  3^x,

u² - 6u + 8

By factorization, we get

(u - 2) (u - 4)

By replacing u  =  3^x, we get

(3^x - 2) (3^x - 4)

So, the solution is (3^x - 2) (3^x - 4)

Example 9 :

25^x + 4(5^x) - 12

Solution :

Given, 25^x + 4(5^x) - 12

=  (5²)^x + 4(5^x) - 12

=  (5^x)² + 4(5^x) - 12

Let u  =  5^x,

=  u² + 4u - 12

Now, we have the quadratic equation form,

u² + 4u - 12

By factorization, we get

(u - 2) (u + 6)

By replacing u  =  5^x, we get

(5^x - 2) (5^x + 6)

So, the solution is (5^x - 2) (5^x + 6)

Example 10 :

64^x + 3(8^x) - 4

Solution :

Given, 64^x + 3(8^x) - 4

=  (8²)^x + 3(8^x) - 4

=  (8^x)² + 3(8^x) - 4

Let u  =  8^x,

=  u² + 3u - 4

Now, we have the quadratic equation form,

u² + 3u - 4

(u - 1) (u + 4)

By replacing u  =  8^x, we get

(8^x - 1) (8^x + 4)

So, the solution is (8^x - 1) (8^x + 4).

Example 11 :

4^x - 2^x - 20

Solution :

Given, 4^x - 2^x - 20

=  (2²)^x - (2^x) - 20

Let u  =  2^x,

=  u² - u - 20

Now, we have the quadratic equation form,

u² - u - 20

(u - 5) (u + 4)

By replacing u  =  2^x, we get

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

So, the solution is (2^x - 5) (2^x + 4).

Example 12 :

25^x + 5^x - 2

Solution :

Given, 25^x + 5^x - 2

=  (5²)^x + (5^x) - 2

Let u  =  5^x,

= u² + u - 2

Now, we have the quadratic equation form,

u² + u - 2

(u + 2) (u - 1)

By replacing u  =  5^x, we get

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

So, the solution is (5^x + 2) (5^x - 1).

Example 13 :

4^x + 9(2)^x + 18

Solution :

Given, 25^x + 5^x - 2

=  (5²)^x + (5^x) - 2

Let u  =  5^x,

= u² + u - 2

Now, we have the quadratic equation form,

u² + u - 2

(u + 2) (u - 1)

By replacing u  =  5^x, we get

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

So, the solution is (5^x + 2) (5^x - 1).

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