# FACTORING EXPONENTIAL EXPRESSIONS USING ALGEBRAIC IDENTITIES

Factoring means, taking common value out. Factoring can be done in the following ways.

i) Grouping

ii) Using algebraic identities

How to do factoring using grouping ?

This can be done by following the steps given below.

Step 1 :

Using the rules of exponents, break up the given exponents.

Step 2 :

Observe the common terms in the expression.

Step 3 :

Take it out and write the leftovers inside the bracket.

How to do factoring using algebraic identities ?

Some times exponential expressions will be in the form or we may have to express it in the above form.

a2 + 2ba + b2

a2 - 2ab + b2

a2 - b2

Then we have to rewrite in the form of (a + b)2, (a - b)2 and (a + b)(a - b) respectively.

Factorise the following :

Problem 1 :

4x - 9

Solution :

= 4x - 9

= (22)x - 32

= (2x)2 - 32

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

= (2+ 3) (2- 3)

Problem 2 :

9x - 25

Solution :

= 9x - 25

= (32)x - 52

= (3x)2 - 52

= (3+ 5) (3- 5)

Problem 3 :

64 - 9x

Solution :

= 64 - 9x

= 8- (3x)2

Looks like a2 - b2

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

= (8 - 3x)(8 + 3x)

Problem 4 :

16 - 25x

Solution :

= 16 - 25x

= 4- (52)x

Exchanging the powers, we get

= 4- (5x)2

= (4 + 5x) (4 - 5x)

Problem 5 :

4x + 6(2x) + 9

Solution :

= 4x + 6(2x) + 9

= (22)x + 6(2x) + 9

Exchanging the powers, we get

= (2x)2 + 6(2x) + 9

Let t = 2x

= 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+ 3)(2x + 3)

Problem 6 :

9x + 10(3x) + 25

Solution :

= 9x + 10(3x) + 25

= (32)x + 10(3x) + 52

Exchanging the powers, we get

= (3x)2 + 10(3x) + 52

Let t = 3x

= 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+ 5)(3x + 5)

Problem 7 :

4x - 14(2x) + 49

Solution :

= 4x - 14(2x) + 49

= (22)x - 14(2x) + 72

Exchanging the powers, we get

= (2x)2 - 14(2x) + 72

Let t = 2x

= 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- 7)(2x - 7)

Problem 8 :

25x - 4(5x) + 4

Solution :

= 25x - 4(5x) + 4

= (52)x - 4(5x) + 22

Exchanging the powers, we get

= (5x)2 - 2(5x)(2) + 22

Let t = 5x

= (5x)2 - 2(5x)(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- 2)(5x - 2)

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