What is Complete Factorization?

Solution:
12x² + 18x³ can be factorized into any of the following ways:
12x² + 18x³ = x (12x +18x²)
12x² + 18x³ = 6x (2x + 3x²)
12x² + 18 x³ = 6x² (2 + 3x)
There may be more ways of expressing the polynomial 12x² + 18x³ as product of two factors. Let us limit to the above three ways.

Which of the above three ways is accepted as standard form of factorization?

The third one.
In the third one, 6x²is the greatest common divisor or highest common factor of the two algebraic expressions (monomials) 12x² and 18x³

Example 1: Factorize x²y² + y²

Solution:
Completely factorize x²y² + y², i.e. write the greatest common factor of x²y² and y² as the common factor.
y² is the highest common factor. To find the other factor, divide each term by the H.C.F. and add the quotients. (x²y²)/y² = x² and y²/y² = 1
Sum of the quotients is x² + 1, which is the other factor. Therefore,
x²y² + y² = y²(x² + 1)

2. Second type of Factorization: By Grouping of Terms

1. Factorize a² + bc + ab + ac
Solution:
let us group terms in a² + bc + ab + ac
But, which terms to group?

Those terms which yield a common factor on grouping!
Group the terms like this:
a² + ab + bc + ac
a is the common factor in a² + ab and b is the common factor in ab + bc
a² + ab = a (a + b) and bc + ac = c (b + a)
now, the common factor is the binomial (a + b).

{note that a + b is same as b + a, i.e., order of addition does not matter, since addition follows closure property which states a + b = b + a}

So, factorization of
a² + ab + bc + ac =
a (a + b) + c (a + b) =
(a + b) (a + c)
(a + b) and (a + c) are two binomial factors into which a² + ab + bc + ac is factorized.

2. Factorize ax + bx + ay + by

Solution:
Group terms with similar literal coefficients and numerical coefficients.
One group is ax + bx, in which x is the same literal coefficient,
and the other is ay + by, in which the y is the same literal coefficient,
In ax + bx, the common factor is x and in ay +by, the common factor is y.
on factorization with common factors, we have:
ax + bx = x ( a + b ) and
ay + by = y (a + b)
so ax + bx + ay + by = x (a + b) + y (a + b).
(a + b) is the common binomial factor for the next step of factorization:
(a + b)(x + y)
Factorization of ax + bx + ay + by = (a + b)(x + y)

3. Factorize p² qx + pq²y + px² y+ qxy²

Solution:
Group p²qx and px²y.
Common factor is px
Factorization of p²qx and px²y = px (pq + xy)
Again, group pq²y and qx y²
common factor is qy
Factorization of pq²y and qxy² = qy (pq + xy)
So we have :
p² qx + pq²y + px² y+ qxy² = px (pq + xy) + qy (pq + xy)
In the next factorization step,
pq + xy is the common binomial factor. We have:
(pq + xy)(px + qy)

Type 3: Factorization of Trinomial 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
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 Can you do the Factorization?

Factorization ofa² + 2ab + b² gives the famous formula:
(a + b) ² = a² + 2ab + b²

Example 1:
Factorize 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
Factorization of
9p² + 24pq + 16q² = (3p + 4q)² = (3p + 4q)(3p + 4q)

Example 2:
Factorize -4x² + 12x + 9
Solution:
In 4x² + 12x – 9, the leading coefficient is –1.
Whenever the leading coefficient is negative, express the given polynomial as follows:
-1(4x² – 12x + 9).
Now, use factorization of perfect trinomial squares method to factorize
4x² – 12x + 9, which is in the form of the well-known algebraic formula:
a² – 2 ab + b² = (a – b) (a – b).
so, 4x² – 12x + 9 = (2x – 3) (2x – 3)
Now, factorization of
4x² + 12x – 9 = -1(4x² – 12x + 9) = -1(2x – 3) (2x – 3)