MULTIPLYING BINOMIALS WITH VARAIBLE EXPONENTS

Expand and simplify :

Problem 2 :

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

Solution :

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

= 2^x ⋅ 2^x + 2^x ⋅ 5 + 1 ⋅ 2^x + 1 ⋅ 5

= 2^2x + 6 ⋅ 2^x + 5

= 4^x + 6 ⋅ 2^x + 5

Problem 3 :

(5^x - 2) (5^x - 7)

Solution :

(5^x - 2) (5^x - 7)

= 5^x ⋅ 5^x - 5^x ⋅ 7 - 2 ⋅ 5^x + 2 ⋅ 7

= 25^x - 9 ⋅ 5^x + 14

Problem 4 :

(2^x + 1)²

Solution :

(2^x + 1)²

= (2^x)² + 2(2^x)(1) + 1²

= 2^2x + 2^{(x+ 1)} + 1

= 4^x + 2^{(x + 1)} + 1

Problem 5 :

(3^x + 2)²

Solution :

(3^x + 2)²

= (3^x)² + 2(3^x)(2) + 2²

= 3^2x + 4(3^x) + 4

= 9^x + 4(3^x) + 4

Problem 6 :

(4^x - 7)²

Solution :

(4^x - 7)²

= (4^x)² - 2(4^x)(7) + 7²

= 4^2x - 14(4^x) + 49

= 16^x - 14(4^x) + 49

Problem 7 :

(3^x + 1)²

Solution :

(3^x + 1)²

= (3^x)² + 2(3^x)(1) + 1²

= 3^2x + 2(3^x) + 1

= 9^x + 2(3^x) + 1

Problem 8 :

(3^x - 8)²

Solution :

(3^x - 8)²

= (3^x)² - 2(3^x)(8) + 8²

= 3^2x - 16(3^x) + 64

= 9^x - 16(3^x) + 64

Problem 9 :

(5^x - 3)²

Solution :

(5^x - 3)²

= (5^x)² - 2(5^x)(3) + 3²

= 5^2x - 6(5^x) + 9

= 25^x - 6(5^x) + 9

Problem 10 :

(x^1/2 + 3) (x^1/2 - 3)

Solution :

(x^1/2 + 3) (x^1/2 - 3)

= x^1/2 ⋅  x^1/2 - x^1/2 ⋅ 3 + 3 ⋅ x^1/2 – 3 ⋅ 3

= x - 9

Problem 11 :

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

Solution :

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

= 2^x ⋅ 2^x - 2^x ⋅ 5 + 5 ⋅ 2^x - 5 ⋅ 5

= 4^x - 25

Problem 12 :

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

Solution :

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

= x^1/2 ⋅ x^1/2 + x^1/2 ⋅ x^-1/2 + x^-1/2 ⋅ x^1/2 + x^-1/2 ⋅ x^-1/2

= x – x^-1

= x – 1/x

Problem 13 :

(x + 3/x)²

Solution :

(x + 3/x)²

= x² + 2(x)(3/x) + (3/x)²

= x² + 6 + 9/x²

Problem 14 :

(e^x – e^-x)²

Solution :

(e^x – e^-x)²

= (e^x)² – 2(e^x)(e^-x) + (e^-x)²

= e^2x – 2 + e^-2x

Problem 15 :

(3 – 2^-x)²

Solution :

(3 – 2^-x)²

= 3² – 2(3)(2^-x) + (2^-x)²

= 9 – 6(2^-x) + 4^-x

= 9 – (6/2^x) + (1/4)^x