FACTORING EXPONENTIAL EXPRESSIONS USING ALGEBRAIC IDENTITIES

Factorise the following :

Problem 1 :

4^x - 9

Solution :

= 4^x - 9

= (2²)^x - 3²

= (2^x)² - 3²

Using the algebraic identities a² - b² = (a + b) (a - b), we get

= (2^x + 3) (2^x - 3)

Problem 2 :

9^x - 25

Solution :

= 9^x - 25

= (3²)^x - 5²

= (3^x)² - 5²

= (3^x + 5) (3^x - 5)

Problem 3 :

64 - 9^x

Solution :

= 64 - 9^x

= 8² - (3^x)²

Looks like a² - b²

After expanding, we will get (a + b) (a - b)

= (8 - 3^x)(8 + 3^x)

Problem 4 :

16 - 25^x

Solution :

= 16 - 25^x

= 4² - (5²)^x

Exchanging the powers, we get

= 4² - (5^x)²

= (4 + 5^x) (4 - 5^x)

Problem 5 :

4^x + 6(2^x) + 9

Solution :

= 4^x + 6(2^x) + 9

= (2²)^x + 6(2^x) + 9

Exchanging the powers, we get

= (2^x)² + 6(2^x) + 9

Let t = 2^x

= t² + 6t + 9

Now it became a quadratic polynomial or trinomial.

= t² + 2 t (3) + 3²

It looks like a² + 2 ab + b²

= (t + 3)² 

= (t + 3)(t + 3)

Applying the value of t, we get

= (2^x + 3)(2^x + 3)

Problem 6 :

9^x + 10(3^x) + 25

Solution :

= 9^x + 10(3^x) + 25

= (3²)^x + 10(3^x) + 5²

Exchanging the powers, we get

= (3^x)² + 10(3^x) + 5²

Let t = 3^x

= t² + 10t + 25

= t² + 2 (5) t + 5²

Looks like a² + 2 ab + b², then use the formula (a + b)²

= (t + 5)²

= (t + 5)(t + 5)

Applying the value of t, we get

= (3^x + 5)(3^x + 5)

Problem 7 :

4^x - 14(2^x) + 49

Solution :

= 4^x - 14(2^x) + 49

= (2²)^x - 14(2^x) + 7²

Exchanging the powers, we get

= (2^x)² - 14(2^x) + 7²

Let t = 2^x

= t² - 14t + 7²

= t² - 2 (7) t + 7²

Looks like a² - 2 ab + b², then use the formula (a - b)²

= (t - 7)²

= (t - 7)(t - 7)

Applying the value of t, we get

= (2^x - 7)(2^x - 7)

Problem 8 :

25^x - 4(5^x) + 4

Solution :

= 25^x - 4(5^x) + 4

= (5²)^x - 4(5^x) + 2²

Exchanging the powers, we get

= (5^x)² - 2(5^x)(2) + 2²

Let t = 5^x

= (5^x)² - 2(5^x)(2) + 2²

= t² - 2 (2) t + 2²

Looks like a² - 2 ab + b², then use the formula (a - b)²

= (t - 2)²

= (t - 2) (t - 2)

Applying the value of t, we get

= (5^x - 2)(5^x - 2)