# FACTORING WITH VARIABLE EXPONENTS

Factories :

Problem 1 :

32x + 3x

Solution :

32x + 3x

= 3x 3x + 3x

= 3x(3x + 1)

Factories :

Problem 2 :

2n + 2 + 2n

Solution :

2n + 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 :

6n + 1 - 6

Solution :

6n + 1 - 6

= 6n 61 - 6

= 6(6n - 1)

Problem 5 :

7n + 2 - 7

Solution :

7n + 2 - 7

= 7n 72 7

= 7(7n + 1 - 1)

Problem 6 :

3n + 2 - 9

Solution :

3n + 2 - 9

= 3n 32 9

= 9(3n – 1)

Problem 7 :

5(2n) + 2n + 2

Solution :

5(2n) + 2n + 2

= 5(2n) + 2n 22

= 5(2n) + 2n 4

= 2n[5 + 4]

= 2n(9)

Problem 8 :

3n + 2 + 3n+1 + 3n

Solution :

3n + 2 + 3n+1 + 3n

= 3n 32 + 3n 31 + 3n

= 3n 9 + 3n 3 + 3n

= 13(3n)

Problem 9 :

2n+1 + 3(2n) + 2n - 1

Solution :

2n+1 + 3(2n) + 2n – 1

= 2n 21 + 3(2n) + 2n 2-1

= 2n(2 + 3 + 1/2)

= 2n(5 + 1/2)

= 2n(11/2)

= (2n 1)11

Problem 10 :

4x + 11(2x) + 18

Solution :

4x + 11(2x) + 18

= (22)x + 11(2x) + 18

= (2x)2 + 11(2x) + 18

2x = t

= t2 + 11t + 18

= (t + 2)(t + 9)

Replacing t by 2x, we get

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

Problem 11 :

4x – 2x - 20

Solution :

4x – 2x - 20

= (22)x – 2x – 20

= (2x)2 – 2x – 20

2x = t

= t2 – t – 20

= (t - 5)(t + 4)

Replacing t by 2x

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

Problem 12 :

9x + 9(3x) + 14

Solution :

9x + 9(3x) + 14

= (32)x + 9(3x) + 14

= (3x)2 + 9(3x) + 14

3x = t

= t2 + 9t + 14

= (t + 2)(t + 7)

Replacing t by 3x

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

Problem 13 :

9x + 4(3x) - 5

Solution :

9x + 4(3x) - 5

= (32)x + 4(3x) – 5

= (3x)2 + 4(3x) – 5

Let 3x = t

= t2 + 4t – 5

= (t - 1)(t + 5)

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

Problem 14 :

25x + 5x - 2

Solution :

= 25x + 5x - 2

= (52)x + 5x – 2

= (5x)2 + 5x – 2

Let 5x = t

= t2 + t – 2

= (t + 2)(t - 1)

= (5x + 2) (5x - 1)

Problem 15 :

49x – 7x + 1 + 12

Solution :

49x – 7x + 1 + 12

= (72)x – 7x + 1 + 12

= (7x)2 – 7x (7) + 12

Let t = 7x

= t2 – 7t + 12

= (t - 4) (t - 3)

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

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