**A note: **Trinomial Perfect Squares have three monomials, in which two terms are perfect squares and one term is the product of the square roots of the two terms which are perfect squares.

## Example

*a2 + 2ab + b2 – *In this trinomial, a2 and b2 are the two perfect squares and 2ab is the product of the square roots of a2 and b2

**Now, can you do the factoring?** Factoring trinomial a2 + 2ab + b2which is a perfect square gives the following famous formula: *(a + b) 2 = a2 + 2ab + b2. *Now, we proceed with a few examples of factoring trinomials of perfect squares.

## Example 1:

Factorize the trinomial *9p2 + 24pq + 16q2*

### Solution:

*9p2 = (3p)2, just like a2*

*16q2 = (4q)2, just like b2*

*24pq = 2(3p)(4q), just like 2ab*

Applying, *(a + b) 2 = a2 + 2ab + b2* for factoring trinomial given above:

*9p2 + 24pq + 16q2 = (3p + 4q)2 = (3p + 4q)(3p + 4q)*

## Example 2:

Factorize trinomial –*4×2 + 12x + 9*

### Solution:

*In ‑4×2+ 12x – 9*, the leading coefficient is –1.

Whenever the leading coefficient is negative, express the given trinomial as follows:

*-1(4×2– 12x + 9).*

Now, use the method of factoring trinomials of perfect squares on *4×2– 12x + 9,*

The trinomial *4×2– 12x + 9* is in the form of the well-known algebraic formula: *a2 – 2ab + b2 = (a – b) (a – b).*

So, *4×2– 12x + 9 = (2x – 3) (2x – 3)*

Therefore, *‑4×2+ 12x – 9 = -1(4×2– 12x + 9) = -1(2x – 3) (2x – 3)*

**Type 4:**

Factoring Trinomials of the form: *x2 + bx + c*

Consider the multiplication or product of *(x + 2) (x + 3)*

*(x + 2) (x + 3) =*

*x.x + x.3 + 2.x + 2.3 =*

*x2 + 3x + 2x + 6 =*

*x2 + 5x + 6*

So, let us discuss below factoring trinomials such as: *x2 + 5x + 6.*

First, understand that:

Factoring trinomial *x2 + 5x + 6* produces two binomial factors of the form *(x +?) (x +?)…………(1), *where the two ? stand for some numbers.

Note that the leading coefficient in the trinomial of the form *x2 + bx + c is 1*. So write 1 for numerical coefficients of x in each binomial factor. Now, what to fill in? in (1) above.

**Follow this Tip:**

Write numbers in each? so that their product is 6, the constant term and sum is 5, the middle term numerical coefficient. Which two numbers’ product is 6 and sum is 5? Did you say it, didn’t you? Yes, it is 2 and 3. Therefore, write 2 and 3 in the? in (1) above. So the two binomial factors are (x + 2) and (x + 3). Therefore, *(x + 2) (x + 3) = x2 + 5x + 6*

## Example 3:

Factorize x2 +13x + 36

### Solution:

The leading coefficient of x2 is 1.

*Set x2 +13x + 36 = (x + a) (x + b), *where a and b are two numbers such that *a + b = 13, and a.b = 36 *to find a and b, express 36 as product of pairs of its factors.

*36 = 36×1, 36 = 2 × 18, 36 = 3 × 12, 36 = 4 × 9 *from factoringof 36 into the above four forms, in the pair 4 and 9, the sum is 13, the numerical coefficient of the middle term in *x2 +13x + 36 *so, in the factoringof *x2 +13x + 36 = (x + a) (x + b) *plug 4 in a and 9 in b {9 in a and 4 in b is equally correct} *x2 +13x + 36 = (x + 4) (x + 9).*

**Type 5:**

Factoringtrinomials of the form *ax2 + bx + c*

We learned Factoring trinomials of *x2 + bx + c* type, in which the coefficient of the leading term is 1.

**How to factorize ax2 + bx + c, in which the coefficient of the leading term ax2 is a, i.e., not 1? **It’s almost the same, with only one slight change.

**“To factorize ax2 + bx + c: **

- think of two numbers whose product is a×c and
- the sum of the two numbers must be b”

## Example 4:

**Factorize:** *3×2 + 12x + 9*

Observe that 3 is the numerical coefficient of the leading term. So it is of the form *ax2 + bx + c.*

Compare the standard form ax2 + bx + c with the given form *4×2 + 13x + 9*

What do you find? You find in the places of a, b and c respectively 4, 13 and 9 a is 4, b is 13 and c is 9.

Now *a.c = 4.9 = 36*

**Think of two terms whose:**

product is 36 (i.e., a.c) and

the sum is 13 (i.e., b, the middle term)

**factorize 36 into pairs of numbers as follows: ***36 = 36×1, 36 = 2 × 18, 36 = 3 × 12, 36 = 4 × 9*

Now, sum of which two factors is 13. They are 4 and 9

Write 4×2 + 13x + 9 as below: *4×2 + 4 x + 9x + 9*

Take out 4x as the common factor in 4×2 + 4x and 9 as the common factor in *9x + 9* and factorize *4×2 + 4 x + 9x + 9 = 4x (x + 1) + 9 (x + 1) = (x + 1) (4x + 9)*

## Example 5:

Factorize* -6×2 +x +1*

### Solution:

In example 2 under type 3, we used the method –1(given polynomial) for factoring when the numerical coefficient of the leading term is negative. Proceed here too in the same way.

*So, (-6×2 +x + 1) = -1(6×2 – x –1)*

*Factorize (6×2 – x –1) *on comparing the given polynomial with the standard form *ax2 + bx + c*, we find: *a = 6, b = –1 and c = –1*

Now, *a × c = 6 × (–1) = –6, and b = –1.*

We need two numbers (factors) whose product is –6 and sum is –1

The numbers are 2 and 3. Put the ‘– ‘for the larger value 3 as sum is –1

*{i.e. – 3 + 2 = – 1}.*

*Also 3 × (-2) = -6*

*6×2 – x –1= (6×2 – 3x + 2x –1)*

*But (-6×2 +x + 1) = -1(6×2 – x –1)= (3x + 1) (2x –1)*

## Example 6:

Factorize *12×2 – x – 1*

### Solution:

*12×2 – x – 1*

Here *a = 12, b = –1 and c = –1*

Now, *a × c = –12 and b = –1*

4 × 3 = 12.

Adjust signs of 4 and 3 so that their sum is -1. So, *–4 + 3 = –1*

Now,

*12×2 – x – 1 =*

*12×2 – 4x + 3x – 1 =*

*4x (3x – 1) + 1(3x – 1) = (4x + 1)(3x – 1)*