Factorization

Basic Algebra > 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.

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.
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)


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