 # Algebra special factoring

Introduction :

In mathematics, algebra special factoring is one important topic. Algebra special factoring is the process of two binomials which have same two terms but it contains opposite terms that separates the terms are calledas conjugates of one other. Factoring is the process of performing multiplication operation in reverse. Let us solve some example problems for algebra special factoring.

## Algebra special cases for factoring

Some special cases of factoring in algebra are

Differences of squares:

(a2 – b2) = (a – b) (a + b)

Perfect Square Formula :

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

Sum of cubes and differences of cubes:

(a3 + b3) = (a + b) (a2 – ab + b2)

(a3 – b3) = (a – b) (a2 + ab + b2)

## Example problems for algebra special factoring:

Different example problems for algebra special factoring are,

Example 1:

Factor the given expression 36x2 – 25

Solution:

Given expression

36x2 – 25

Using differences of squares in algebra special factoring

(a2 – b2) = (a – b) (a + b)

Using the above the algebra special cases of factoring

36x2 – 25 = (62x2– 52)

= (6x – 5) (6x + 5)

= 6x – 5

Simplify the term

6x – 5 = 0

6x = 5

X = 5/ 6

Solution to the given expression is x = 5/ 6.

Example 2:

Factor the given polynomial expression

25x2 + 1 + 10x

Solution:

Given expression in the form of (a + b)2 = a2 + 2ab + b2

Step 1:

Given polynomial expression

25x2 + 1 + 10x

Step 2:

Given polynomial is in the standard form ax² + bx + c.

25x2 + 10x + 1

Here a = 25, b = 10 and c = 1

Step 3:

Multiply the coefficients of a and c.

25 × 1 = 25

Step 4:

Different ways for get the product of ac, we multiply two integers.

1 × 25 = 12

5 × 5 = 25

Step 5:

If we add a and c then we get the result as equals to b means choose that way.

5 + 5 = 10 is equal to b

Step 6:

Factor the expression

25x2 + 5x + 5x + 1 = 0

5x (5x + 1) + 1 (5x + 1) = 0

(5x + 1) (5x + 1) = 0

Step 7:

Solve the terms.

5x + 1 = 0

5x = -1

x="-1/" 5

Solution: x = -1/ 5