**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.

Some special cases of factoring in algebra are

**Differences of squares:**

(a^{2} – b^{2}) = (a – b) (a + b)

**Perfect Square Formula :**

(a + b)^{2} = a^{2} + 2ab + b^{2}

(a – b)^{2} = a^{2} – 2ab + b^{2}

**Sum of cubes and differences of cubes:**

(a^{3} + b^{3}) = (a + b) (a^{2} – ab + b^{2})

(a^{3} – b^{3}) = (a – b) (a^{2} + ab + b^{2})

Different example problems for algebra special factoring are,

**Example 1:**

Factor the given expression 36x^{2} – 25

**Solution:**

Given expression

36x^{2} – 25

Using differences of squares in algebra special factoring

(a^{2} – b^{2}) = (a – b) (a + b)

Using the above the algebra special cases of factoring

36x^{2} – 25 = (6^{2}x^{2}– 5^{2})

= (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

25x^{2} + 1 + 10x

**Solution:**

** **Given expression in the form of** (**a + b)^{2} = a^{2} + 2ab + b^{2}

**Step 1**:

Given polynomial expression

25x^{2} + 1 + 10x

**Step 2:**

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

25x^{2} + 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

25x^{2 }+ 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