﻿ Factorization

## Factorization

### Type 4: Factorization of 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 =
x 2 + 3x + 2x + 6 =
x 2 + 5x + 6

So, if you were asked to factorize x 2 + 5x + 6, how would you do it?
First, understand that:

factorization of x 2 + 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 x 2 + bx + c is 1.

So write 1 for numerical coefficients of x in each binomial factor.
Now, what to fill in? in (1) above.

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.
Set x 2 +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 factorization of 36 into the above four forms, in the pair 4 and 9, the sum is 13, the numerical coefficient of the middle term in x 2 +13x + 36

so, in the factorization of x 2 +13x + 36 = (x + a) (x + b)
plug 4 in a and 9 in b {9 in a and 4 in b is equally correct}
x 2 +13x + 36 = (x + 4) (x + 9)

### Type 5: Factorization of trinomials of the form: ax 2 + bx + c

We learnt Factorization of x 2 + bx + c in the last 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 ax2 is a, i.e., not 1?

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 ax 2 + bx + c with the given form 4x2 + 13x + 9

What do you find?
You find in the places ofa, 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)

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 4x 2 + 13x + 9 as below:
4x 2 + 4 x + 9x + 9
Take out 4x as the common factor in 4x 2 + 4x
and 9 as the common factor in 9x + 9 and factorize
4x 2 + 4 x + 9x + 9
= 4x (x + 1) + 9 (x + 1)
= (x + 1) (4x + 9)

Example 2:
Factorize -6x 2 +x +1

Solution:
In example 2 under Type 3 , we used the method –1(given polynomial) for factorization when 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 (6x 2 – x –1)

on comparing the given polynomial with the standard form ax 2 + 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

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

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)

### Type 6: Factorization of Difference of Two Perfect Squares

The polynomials (algebraic expressions) are two perfect squares separated by a “negative sign”.

Hence the algebraic expression is called "Difference of Two Perfect Squares".
And this type is called factorization of "Difference of Two Perfect Squares"

Let one perfect square be:
a 2 and the other be b 2
then the difference of the two perfect squares looks:
a 2 – b 2

you know very well that
a 2 – b 2 = (a – b) (a + b)
In (a – b) (a + b):
a is the square root of a 2 and b is the square root of b 2
So, to factorize:

• find square roots of each perfect square.
• In one factor, write their sum and in the other, write their difference.

Example:
Factorize 9x 2 – 16y 2

Solution:
9x 2 = (3x) 2 and 16 y 2 = (4y) 2

Using the algebraic formula:
a 2 – b 2 = (a – b) (a + b)
factorization of
9x 2 – 16y 2 = (3x – 4y) (3x + 4y)

{order of factors does not matter, so it is equally correct to write the product as (3x + 4y) (3x – 4y)}

Example 2
Factorize (p2 / 4) – (q2 / 9)

Solution:
first find square roots of the two perfect squares.
(p2 / 4) = (p/2)2 and (q/2) 2
Next, use the algebraic formula
a 2 – b 2 = (a – b) (a + b)
to factorize(p2 / 4) – (q2 / 9) =
(p/2 – q/2) (p/2 + q/2)

Example 3:
Factorize 27y 2 – 48z 2

Solution:
3 is a common factor of 27 and48.
Write it outside the bracket as follows:
3 (9y 2 – 16z 2)

Now use the algebraic formula:
a 2 – b 2 = (a – b) (a + b)
to factorize 9y 2 – 16z 2
9y 2 – 16z 2 = (3y) 2 – (4z) 2
9y 2 – 16z 2 = (3y – 4z) (3y + 4z)
Therefore, 3 (9y 2 – 16z 2) = 3 (3y – 4z) (3y + 4z)

Example 4:
Factorize 4x 2 – 25(y + z) 2

Solution:
4x 2 = (2x) 2 and 25 (y + z) 2 = [5 (y + z)] 2
4x 2 – 25(y + z) 2 = (2x) 2 – [5 (y + z)] 2
= (2x –5(y + z)) (2x + 5(y + z)) =

Now take the common factor 5 inside the bracket and write it as numerical coefficient for each of y and z
(2x – 5y – 5z) (2x + 5y + 5z)