__A note before factoring trinomials which are perfect squares: __

*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

a^{2} + 2ab + b^{2}

In this trinomial, a^{2} and b^{2} are the two perfect squares and 2ab is the product of the square roots of a^{2} and b^{2}

Now, can you do the factoring?

Factoringtrinomial a^{2} + 2ab + b^{2}which is a perfect square gives the following famous formula:

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

Now, we proceed with a few examples on factoring trinomials of perfect squares

**Example 1: **

**Factorize the trinomial 9p ^{2} + 24pq + 16q^{2}**

**Solution: **

9p^{2} = (3p)^{2}, just like a^{2}

16q^{2} = (4q)^{2}, just like b^{2}

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

Applying, (a + b) 2 = a^{2} + 2ab + b^{2} for factoring trinomial given above:

9p^{2} + 24pq + 16q^{2} = (3p + 4q)^{2} = (3p + 4q)(3p + 4q)

**Example 2: **

Factorize trinomial -4x^{2} + 12x + 9

**Solution: **

In ‑4x^{2}+ 12x – 9, the leading coefficient is –1.

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

-1(4x

Now, use the method of factoringtrinomialsof perfect squares on

4x2– 12x + 9,

The trinomial 4x

a

So, 4x

Therefore,

‑4x

Type 4:

**FactoringTrinomials of the form: x ^{2} + bx + c**

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

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

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

*x ^{2} + 3x + 2x + 6 = *

So, let us discuss below factoring trinomials such as:

First, understand that:

*Note that the leading coefficient in the trinomial of the form x ^{2} + bx + c is 1. *

*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?*

*You said 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) = x ^{2} + 5x + 6*

**Example 2: **

**Factorize x^{2} +13x + 36**

**Solution: **

Leading coefficient of *x ^{2} is 1. *

Where

to find a and b, express 36 as product of pairs of its factors.

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

so, in the factoringof

plug 4 in a and 9 in b {9 in a and 4 in b is equally correct}

Type 5:

**Factoringtrinomials of the form ax ^{2} + bx + c**

We learnt Factoring trinomials of x^{2} + bx + c type, in which the coefficient of the leading term is 1.

How to factorize **ax ^{2} + bx + c, **in which the coefficient of the leading term ax

Its almost the same, with only one slight change.

To factorize ax^{2} + bx + c:1. think of two numbers whose product is a×c and 2. the sum of the two numbers must be b |

**Example 1:**

Factorize: 3x^{2} + 12x + 9

Observe that 3 is the numerical coefficient of the leading term. So it is of the form

**ax ^{2} + bx + c.**

Compare the standard form ax2 + bx + c with the given form 4x

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

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

4x

Take out 4x as the common factor in

and 9 as the common factor in

4x

**Example 2: **

**Factorize -6x ^{2} +x +1**

**Solution: **

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

So, **(-6x ^{2} +x + 1) = -1(6x^{2} – x –1)**

Factorize

on comparing the given polynomial with the standard form

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

6x

= 3x (2x –1) + 1(2x – 1)

= (3x + 1) (2x –1)

But (-6x

**Example 3:**

**Factorize 12x ^{2} – x – 1 **

**Solution: **

12x^{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,

12x^{2} – x – 1 =

12x^{2} – 4x + 3x – 1 =

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