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FACTORIZATION

FUNDAMENTALS

FACTORS: When an algebraic quantity can be expressed as the product of two or more algebraic quantities, then each of these quantities is called a factor of the given algebraic quantity and the process of finding factors, is called FACTORIZATION.

Remarks: Factorization is the opposite process of multiplication,

EXAMPLE Look at the examples given below:

• It is advisable that students memorize squares of numbers from 1 to 20. E.g. here, \[\mathbf{1}{{\mathbf{5}}^{\mathbf{2}}}=\mathbf{225}\] and \[\mathbf{1}{{\mathbf{7}}^{\mathbf{2}}}=\mathbf{289}\] are readily used.

1. Factorization when a Common Monomial Factor Occurs in Each Term.

METHOD: Step 1. Find the HCF of all the terms.

Step2. Divide each term by this HCF.

Step3. Write the given expression= HCF \[\times \] quotient obtained in step 2.

Conceptual Framework / Idea behind above steps: HCF itself is one of the factors. Hence,

\[\text{other}\,\text{factor}\,\text{will}\,\text{be}\,\text{equal}\,\text{to}\frac{\text{Given}\,\text{pression}}{\text{HCF}}\]

EXAMPLE 1. Factorize (i.e. break into factors) each of the following:

(1) \[13n+117\]             

(2) \[{{n}^{3}}+2n+{{n}^{2}}\]        

(3) \[15{{x}^{2}}{{y}^{2}}{{z}^{2}}+5x{{y}^{2}}z+5xyz\]

(4) \[6ab-9bc\]

(1) \[13n+117=13\left( n+9 \right)\]

(2) \[{{n}^{3}}+2n+{{n}^{2}}=n({{n}^{2}}+2+n)\] 

(3) \[15{{x}^{2}}{{y}^{2}}{{z}^{2}}+5x{{y}^{2}}z+5xyz=5xyz(3xyz+y+1)\]

2. Factorization when one or more Binomial is Common

METHOD: Step 1. Find the common binomial by intelligent thinking or by trial & error.

Step 2. Divide each term by this common binomial.

Step 3. Write the given expression = this binomial \[\times \] quotient obtained in Step 2

EXAMPLES. Factorize:

(1) \[6x\left( 3a-4b \right)+10y\left( 3a-4b \right)\]     

(2) \[6\left( 16x-23y \right)-22{{\left( 16x-23y \right)}^{2}}\]

(3) \[mn{{\left( ax-2by \right)}^{2}}+m{{n}^{2}}(ax-2by)\]

We have,

(1) \[6x\left( 3a-4b \right)+10y\left( 3a-4b \right)\]

\[=\left( 6x+10y \right)\left( 3a-4b \right)\]

(2) \[6\left( 16x-23y \right)-22{{\left( 16x-23y \right)}^{2}}\]

\[=\left( 16x-23y \right)\left[ 6-22\left( 16x-23y \right) \right]\]

\[=\left( 16x-23y \right)\times \left( -346x+506y \right)\]

\[=2\text{ }x\left( 16x-23y \right)x\left( -173x+253y \right)\]

(3) \[mn{{\left( ax-2by \right)}^{2}}+m{{n}^{2}}(ax-2by)\]

\[=mn\left( ax-2by \right)\times (ax-2by+n)\]

3. Factorization by Grouping

The terms of the given expression are arranged in suitable groups so that all the groups have a common factor. The key idea is (1) to identify the common factor (2) take out this common factor.

EXAMPLE 4. Factorize:

(1) \[{{m}^{2}}+np+mn+mp\]           

(2) \[a{{x}^{2}}+b{{y}^{2}}+a{{y}^{2}}+b{{x}^{2}}\]  

(3) \[1-a-b+ab\]

(4) \[xy-ny+mn-mx\]

(5) \[1+y+yz+{{y}^{2}}z\]       

(6) \[xy\left( {{m}^{2}}+{{n}^{2}} \right)+mn\left( {{x}^{2}}+{{y}^{2}} \right)\]

Solution: By suitably rearranging the terms, we have:

(1) \[{{m}^{2}}+np+mn+mp={{m}^{2}}+mn+np+mp\]

\[=m(m+n)+p(m+n)=(m+n)(m+p).\]

(2) \[a{{x}^{2}}+b{{y}^{2}}+a{{y}^{2}}+b{{x}^{2}}=a{{x}^{2}}+b{{x}^{2}}+b{{y}^{2}}+a{{y}^{2}}\]\[={{x}^{2}}\left( a+b \right)+{{y}^{2}}\left( b+a \right)=\left( {{x}^{2}}+{{y}^{2}} \right)\left( a+b \right).\]

(3) \[1-a-b+ab=1-a-b\left( 1-a \right)\]

\[~=1\left( 1-a \right)-b\left( 1-a \right)=\left( 1-a \right)\left( 1-b \right)\]

(4) \[xy-ny+mn-mx=\left( x-n \right)y+m\left( n-x \right)\]

\[=\left( x-n \right)y-m\left( x-n \right)\left( x-n \right)\left( y-m \right)\]

(5) \[1+y+yz+y=(1+y)+yz(1+y)\]

\[=1\times (1+y)+yz\times (1+y)=(1+y)\times \left( 1+yz \right).\]

(6) \[xy\left( {{m}^{2}}+{{n}^{2}} \right)+mn\left( {{x}^{2}}\text{+ }{{y}^{2}} \right)\]

\[=xy{{m}^{2}}+xy{{n}^{2}}+mn{{x}^{2}}+mn{{y}^{2}}\]

Consider factors "mx" common between 1^st & 3^rd terms. Similarly, consider factors "ny" common between 2^nd & 4^th terms.

\[\Rightarrow mx\left( my+nx \right)+ny\left( nx+my \right)\]\[=\left( mx+ny \right)\times \left( nx+my \right)\]

4. Factorization when given term is a Perfect Square

FORMULA:

(i) \[{{\mathbf{a}}^{\mathbf{2}}}+{{\mathbf{b}}^{\mathbf{2}}}+2\mathbf{ab}={{\left( \mathbf{a}+\mathbf{b} \right)}^{\mathbf{2}}}\]

(ii) \[{{\mathbf{a}}^{\mathbf{2}}}+{{\mathbf{b}}^{\mathbf{2}}}-\mathbf{2ab}={{\left( \mathbf{a}-\mathbf{b} \right)}^{\mathbf{2}}}\]

(1) \[~{{x}^{2}}+20x+100\]  

(2) \[{{a}^{2}}{{x}^{2}}+2\text{ }abxy+{{b}^{2}}{{y}^{2}}\]

(3) \[{{y}^{2}}-26xy+169\]                        

(4) \[~{{y}^{2}}-6my+9{{m}^{2}}\]

Let us illustrate through one example. Rest you should try yourself. Let us consider example

(2) \[{{a}^{2}}{{x}^{2}}+2abxy+{{b}^{2}}+{{y}^{2}}\]

\[={{(ax)}^{2}}+2\times (ax)\times (by)+{{(by)}^{2}}\]

It is of the form \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( ax+by \right)}^{2}}\]

5. Factorization when given term is Difference of Two Squares

FORMULA: \[\left( {{\mathbf{a}}^{\mathbf{2}}}-{{\mathbf{b}}^{\mathbf{2}}} \right)=\left( \mathbf{a}+\mathbf{b} \right)\left( \mathbf{a}-\mathbf{b} \right)\]

(1) \[81-{{x}^{2}}\]                             

(2) \[{{m}^{2}}{{\text{y}}^{2}}-81{{n}^{2}}\]

Let us illustrate through one example. Let us consider example: (2) \[{{m}^{2}}{{y}^{2}}-81{{n}^{2}}\]

                            \[={{(my)}^{2}}-{{(9n)}^{2}}={{(my-9n)}^{2}}\]

6. Factorization of Quadratic Trinomials

When trinomial is of the form: \[\left( {{\mathbf{x}}^{\mathbf{2}}}+\mathbf{mx}+\mathbf{n} \right)\]

For factorizing\[\left( {{x}^{2}}+mx+n \right),\]we find two numbers a and b such that \[(a+b)=m\] and ab = n. Then,

\[{{x}^{2}}+mx+n={{x}^{2}}+(a+b)x+ab=(x+a)(x+b).\]

This is essentially based on concepts discussed under quadratic equations in GMO, Class VII book. Basically , we need to find roots of quadratic equation; a and b are roots which may be found by HIT & TRIAL as discussed above or by exact method. Roots \[=\frac{-m\pm \sqrt{{{m}^{2}}-4n}}{2}\]

E.g. Factorize: \[{{x}^{2}}+13x+42\]

Solution:  In the given expression, sum of roots =13 and product of roots =42.

Clearly, the numbers are 6 and 7.

\[\therefore {{x}^{2}}+13x+42={{x}^{2}}+6x+7x+6\times 7\]

\[=x\left( x+6 \right)+7\left( x+6 \right)=\left( x+6 \right)\left( x+7 \right).\]

When trinomial is of the form: \[\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{bx}+\mathbf{c}\]

For factorizing \[\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{bx}+\mathbf{c}\]we split b into two parts whose sum is b and product is ac. Then, proceed to factorize.

E.g. \[6{{x}^{2}}+7x+2\]

Solution:  In the given expression, \[6{{x}^{2}}+7x+2\], we have to find two numbers whose sum is 7 and product \[=ac=6\times 2=12.\]

The numbers are 3 and 4.

\[\therefore 6{{x}^{2}}+7x+2=6{{x}^{2}}+3x+4x+2\]

\[=3x\left( 2x+1 \right)+2\left( 2x+1 \right)=\left( 3x+2 \right)\left( 2x+1 \right).\]