Factorization: When an algebraic expression can be written as a product of two or more expressions, then each of these expressions is called a factor of the given expression. And the process is called factorization.

Example:

$(x^2-25)=(x-5)(x+5)$   or

$x^2+9+6x=(x+3)(x+3)$

H.C.F of Monomials

H.C.F of Monomials $= ($ H.C.F. of numerical coefficient $) \times ($ H.C.F. of Literal coefficient $)$

Example:

$6xy^3$ and $9x^2 y^2 = ($ H.C.F of $6$ and $9) \times ($ H.C.F. of  $xy^3$  and $x^2 y^2 )= 3xy^2$

Factorization of an expression by taking out the common factor

Case 1

When the expression is in the form of $ax+by$   then proceed as follows:

Step 1: Find the HCF of all the terms of the expression

Step 2: Divide each of the terms with the HCF obtained in step 1

Let’s do an example. Factorize $6x^2-8xy+4x$

Step 1: Find the HCF of $6x^2, \ 8xy, \ 4x \$ which  is $2x$

Step 2: Therefore, $2x$ is common in all the terms.

Hence,$6x^2-8xy+4x=2x(6x-4y+2)$

Case 2

In case if a polynomial is a common multiplier of each term of the given expression, then first take the common multiplier and then use distributive law.

Expression would look something like this…$a(x+y) \pm b(x+y)$. In this case $(x+y)$  is common and we could take that out

Let’s do one example for Case 2. Factorize:

$3a(x+2y)-2b(x+27)=(x+2y)(3a-2b)$

Factorization of an expression by Grouping the Terms

The expression of the form $ac + bd + ad + bc = a(c + d) + b(c + d) = (a + b)(c + d)$

Factorizing the difference of two squares

Algebraic expressions like $(a^2-b^2 )=(a+b)(a-b)$

Factorization of perfect square trinomials

The algebraic expressions of the form

$a^2+b^2 + 2ab \ or \ a^2+b^2-2ab$   can be factorize using the formula $(a+b)^2=a^2+b^2+2ab$  or $(a-b)^2=a^2+b^2-2ab$

Example: $x^2+14x+49= (x+7)^2$

Factorization of Trinomials of the form $Ax^2+Bx+C$

In such a case, find two numbers  $a$ and $b$ such that $a+b=B$  and $ab=AC$

Let’s do an example for this as well.

Factorize,   $3x^2+11x+10$

Let $a$   and $b$ be two numbers

Therefore $a + b = 11$   and   $ab = 30$. Hence calculating for $a$ and $b$ we get $6$ and $5$. Therefore

$3x^2+11x+10$

$=3x^2+6x+5x+10$

$=3x(x+2)+5(x+2)$

$=(3x+5)(x+2)$