\displaystyle \textbf{FACTORIZATION}

\displaystyle \text{When an algebraic expression is written as a product of two or more}
\displaystyle \text{expressions, each of these expressions is called a factor of the}
\displaystyle \text{given expression. The process of expressing an algebraic expression}
\displaystyle \text{as a product of its factors is called factorization.}

\displaystyle \text{For example:}
\displaystyle x^2-25=(x-5)(x+5).
\displaystyle x^2+6x+9=(x+3)(x+3)=(x+3)^2.
\displaystyle \\

\displaystyle \textbf{H.C.F. OF MONOMIALS}

\displaystyle \text{The H.C.F. of two or more monomials is the product of the H.C.F.}
\displaystyle \text{of their numerical coefficients and the H.C.F. of their literal parts.}

\displaystyle \text{H.C.F. of monomials} =(\text{H.C.F. of numerical coefficients})
\displaystyle \times(\text{H.C.F. of literal parts}).

\displaystyle \text{For the literal part, take each common variable with the smallest}
\displaystyle \text{power in which it occurs in the given monomials.}

\displaystyle \text{Example: Find the H.C.F. of }6xy^3\text{ and }9x^2y^2.
\displaystyle \text{H.C.F. of }6\text{ and }9=3.
\displaystyle \text{H.C.F. of }xy^3\text{ and }x^2y^2=xy^2.
\displaystyle \therefore \text{H.C.F.}=3xy^2.
\displaystyle \\

\displaystyle \textbf{FACTORIZATION BY TAKING OUT THE COMMON FACTOR}

\displaystyle \textbf{Case 1: Common Monomial Factor}

\displaystyle \text{When all the terms of an algebraic expression have a common factor,}
\displaystyle \text{the expression can be factorized by taking out its H.C.F.}

\displaystyle \text{Step 1: Find the H.C.F. of all the terms of the expression.}
\displaystyle \text{Step 2: Divide each term by the H.C.F.}
\displaystyle \text{Step 3: Write the H.C.F. outside the brackets and the quotients inside.}

\displaystyle \text{Example: Factorize }6x^2-8xy+4x.
\displaystyle \text{The H.C.F. of }6x^2,\ 8xy\text{ and }4x\text{ is }2x.
\displaystyle \therefore 6x^2-8xy+4x
\displaystyle =2x(3x-4y+2).
\displaystyle \\

\displaystyle \textbf{Case 2: Common Polynomial Factor}

\displaystyle \text{If the same polynomial occurs as a factor in every term, it can be}
\displaystyle \text{taken out as a common factor using the distributive law.}

\displaystyle a(x+y)\pm b(x+y)=(x+y)(a\pm b).

\displaystyle \text{Example: Factorize }3a(x+2y)-2b(x+2y).
\displaystyle 3a(x+2y)-2b(x+2y)
\displaystyle =(x+2y)(3a-2b).
\displaystyle \\

\displaystyle \textbf{FACTORIZATION BY GROUPING THE TERMS}

\displaystyle \text{Some expressions can be factorized by grouping their terms so that}
\displaystyle \text{each group has a common factor.}

\displaystyle ac+bd+ad+bc
\displaystyle =(ac+ad)+(bc+bd)
\displaystyle =a(c+d)+b(c+d)
\displaystyle =(a+b)(c+d).

\displaystyle \text{Example: Factorize }ax+ay+bx+by.
\displaystyle ax+ay+bx+by
\displaystyle =a(x+y)+b(x+y)
\displaystyle =(a+b)(x+y).
\displaystyle \\

\displaystyle \textbf{FACTORIZING THE DIFFERENCE OF TWO SQUARES}

\displaystyle \text{The difference of two squares can be factorized using the identity:}
\displaystyle a^2-b^2=(a+b)(a-b).

\displaystyle \text{Example: Factorize }x^2-25.
\displaystyle x^2-25=x^2-5^2
\displaystyle =(x+5)(x-5).
\displaystyle \\

\displaystyle \textbf{FACTORIZATION OF PERFECT SQUARE TRINOMIALS}

\displaystyle \text{A trinomial is a perfect square if it is of either of the forms:}
\displaystyle a^2+2ab+b^2=(a+b)^2,
\displaystyle a^2-2ab+b^2=(a-b)^2.

\displaystyle \text{Thus, the first and last terms must be perfect squares, and the}
\displaystyle \text{middle term must be twice the product of their square roots.}

\displaystyle \text{Example: Factorize }x^2+14x+49.
\displaystyle x^2+14x+49
\displaystyle =x^2+2(x)(7)+7^2
\displaystyle =(x+7)^2.

\displaystyle \text{Similarly, }x^2-10x+25
\displaystyle =x^2-2(x)(5)+5^2
\displaystyle =(x-5)^2.
\displaystyle \\

\displaystyle \textbf{FACTORIZATION OF TRINOMIALS OF THE FORM }Ax^2+Bx+C

\displaystyle \text{To factorize }Ax^2+Bx+C,\text{ split the middle term }Bx\text{ into}
\displaystyle \text{two terms whose coefficients have sum }B\text{ and product }AC.

\displaystyle \text{That is, find two numbers }p\text{ and }q\text{ such that}
\displaystyle p+q=B\qquad\text{and}\qquad pq=AC.

\displaystyle \text{Then split }Bx\text{ as }px+qx\text{ and factorize by grouping.}

\displaystyle \text{Example: Factorize }3x^2+11x+10.
\displaystyle A=3,\qquad B=11,\qquad C=10.
\displaystyle AC=3(10)=30.
\displaystyle \text{We need two numbers whose sum is }11\text{ and product is }30.
\displaystyle 6+5=11\qquad\text{and}\qquad 6(5)=30.
\displaystyle \therefore 3x^2+11x+10
\displaystyle =3x^2+6x+5x+10
\displaystyle =3x(x+2)+5(x+2)
\displaystyle =(3x+5)(x+2).
\displaystyle \\

\displaystyle \textbf{FACTORIZATION OF THE SUM AND DIFFERENCE OF TWO CUBES}

\displaystyle \text{The following identities can also be used for factorization:}
\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2),
\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2).

\displaystyle \text{Example: Factorize }x^3-27.
\displaystyle x^3-27=x^3-3^3
\displaystyle =(x-3)(x^2+3x+9).
\displaystyle \\

\displaystyle \textbf{CHECKING A FACTORIZATION}

\displaystyle \text{A factorization can be checked by multiplying the factors obtained.}
\displaystyle \text{If their product is equal to the original expression, the}
\displaystyle \text{factorization is correct.}

\displaystyle \text{For example:}
\displaystyle (3x+5)(x+2)
\displaystyle =3x^2+6x+5x+10
\displaystyle =3x^2+11x+10.
\displaystyle \\

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