Factorization

Category : 9th Class

*         Factorization

 

The following are the different methods for factorization of a polynomials.  

 

*            Taking Out Common Factor

 

If every term of a polynomial has a common factor then we divide each term by the common factor and multiply it by the remaining.  

 

 

Factories: \[7{{(a+b)}^{2}}-14(a+b)\]        

(a)\[(a+b)]7(a+b)-14]\]                

(b) \[7[{{(a+b)}^{2}}-2(a+b)]\] (c) \[7(a+b)(a+b-2)\]                     

(d) \[7(a+b)(a+b-2a)\]

(e) None of these  

 

Answer: (c)  

Explanation:

Here, 7 and \[(a+b)\] is a common factor. Therefore, \[7{{(a+b)}^{2}}-14(a+b)=7(a+b)(a+b-2)\]  

 

*            By Grouping

 

In this method, we group a given expression in such a way that we have a common factor. This method is effective for the expression in which it is not possible to take out common factor directly.  

 

 

Factories:  \[{{a}^{2}}{{y}^{2}}+(a{{y}^{2}}+1)y+a\]

(a) \[({{a}^{2}}y+1)(ay+1)\]                        

(b) \[(ay+y)({{a}^{3}}y+1)\]

(c) \[({{a}^{2}}y+1)(a+y)\]                           

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

(e) None of these  

 

Answer: (c)  

Explanation: 

\[{{a}^{2}}{{y}^{2}}+(a{{y}^{2}}+1)y+a={{a}^{2}}{{y}^{2}}+a{{y}^{3}}+y+a=\]\[{{a}^{2}}{{y}^{2}}+a+a{{y}^{3}}+y\] \[=a(a{{y}^{2}}+1)+y(a{{y}^{2}}+1)+(a+y)\]    

 

*            By Making a Trinomial of a Perfect Square                     

Adjust the trinomial in such a way that there must be a common factor.  

 

 

The factor of \[{{\left( 3x-\frac{1}{y} \right)}^{2}}-8\left( 3x-\frac{1}{y} \right)+16+\left( z+\frac{1}{y}-2x \right)\left( 3x-\frac{1}{y}-4 \right)\]

(a)\[\left( 3x-\frac{1}{y}-4 \right)(x+z-4)\]           

(b) \[\left( 3x-\frac{1}{y}-4 \right)(3x+z-4)\]

(c) \[\left( 3x-\frac{1}{y}-4 \right)(2x+3z-4)\]     

(d) \[\left( 3x-\frac{1}{y}-4 \right)(x+3z-4)\]

(e) None of these  

 

Answer: (a)  

Explanation:

\[{{\left( 3x-\frac{1}{y} \right)}^{2}}-8\left( 3x-\frac{1}{y} \right)+16+\left( z+\frac{1}{y}-2x \right)\left( 3x-\frac{1}{y}-4 \right)\]

\[={{\left( 3x-\frac{1}{y} \right)}^{2}}-2\times 4\left( 3x-\frac{1}{y} \right)+{{4}^{2}}+\left( z+\frac{1}{y}-2x \right)\left( 3x-\frac{1}{y}-4 \right)\]

\[={{\left( 3x-\frac{1}{y}-4 \right)}^{2}}+\left( z+\frac{1}{y}-2x \right)+\left( 3x-\frac{1}{y}-4 \right)\]

\[=\left( 3x-\frac{1}{y}-4 \right)\left[ 3x-\frac{\bcancel{1}}{\bcancel{y}}-4+z+\frac{\bcancel{1}}{\bcancel{y}}-2x \right]\]

\[=\left( 3x-\frac{1}{y}-4 \right)(x-4+z)\]  

 

*            By Using Different Formulae

By the inspection of given expression use suitable formula (Identifies) and factories it.  

 

 

 

The factor of \[{{y}^{6}}+4{{y}^{2}}-1\] is

(a) \[({{y}^{2}}+y-1)({{y}^{4}}-{{y}^{3}}+2{{y}^{2}}+y+1)\]

(b) \[({{y}^{2}}+y-1)({{y}^{4}}-{{y}^{3}}+y+1)\]

(c) \[(y+{{y}^{3}}-1)({{y}^{4}}-{{y}^{3}}+{{y}^{2}}+y+1)\]

(d) \[({{y}^{2}}+y-1)({{y}^{4}}-{{y}^{3}}+{{y}^{2}}+y-1)\]

(e) None of these  

 

Answer: (a)  

Explanation:

\[{{y}^{6}}+4{{y}^{3}}-1={{y}^{6}}+{{y}^{3}}+3{{y}^{3}}-1)={{y}^{6}}+{{y}^{3}}+{{(-1)}^{3}}+3{{y}^{3}}\]

\[={{({{y}^{2}})}^{3}}+({{y}^{3}})+{{(-1)}^{3}}-3.({{y}^{2}}).y(-1)\]

\[=({{y}^{2}}+y-1)[{{({{y}^{2}})}^{2}}+{{y}^{2}}+{{(-1)}^{2}}-{{y}^{2}}.y-y(-1)-(-1)\times {{y}^{2}}\]

[Using Formula:\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)\] \[({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)]\]

\[=({{y}^{2}}+y-1)({{y}^{4}}+{{y}^{2}}-1-{{y}^{3}}+y+{{y}^{2}})\] \[=({{y}^{2}}+y-1)({{y}^{4}}+{{y}^{3}}+2{{y}^{2}}+y-1)\]  

 

*            Factorization of Quadratic Polynomial

 

The general form of a quadratic polynomial is \[a{{x}^{2}}+bx+c\], where \[a\ne 0\], such expressions are factories as following:  

Step 1: Take "a" common from whole expression, i.e.\[a\left( {{x}^{2}}+\frac{b}{a}x+\frac{c}{a} \right)\]  

Step 2: Factories the expression \[a\left( {{x}^{2}}+\frac{b}{a}x+\frac{c}{a} \right)\] by converting it as the different of two squares.  

 

 

 

The factor of \[-10{{a}^{2}}+31a-24\] is______.

(a) \[\text{(2a}-\text{3})(\text{5a}-\text{8)}\]                  

(b) \[\text{(3}-\text{2a})(\text{8}-\text{5a)}\]

(c) \[\text{(2a}+\text{3})(\text{5a}-\text{8})\]                  

(d) \[\text{-(2a}-\text{3})(\text{5a}-\text{8})\]

(e) None of these  

 

Answer: (d)  

Explanation:  

The given expression is \[-\text{1}0{{a}^{\text{2}}}+\text{31a}-\text{24}\]  

Step 1: Taking -10, common from whole expression then the expression will be\[-10\left( {{a}^{2}}-\frac{31}{10}a+\frac{24}{10} \right)\]  

Step 2:  \[-10\left( {{a}^{2}}-\frac{31}{10}a+\frac{24}{10} \right)\]

\[=-10\left[ {{a}^{2}}-2.\frac{31}{20}a+{{\left( \frac{31}{20} \right)}^{2}}-{{\left( \frac{31}{20} \right)}^{2}}+\frac{24}{10} \right]\]

\[=-10\left[ {{a}^{2}}-2.\left( \frac{31}{20} \right)a+{{\left( \frac{31}{20} \right)}^{2}}-\frac{961-960}{400} \right]=\]\[-10\left[ {{\left( a-\frac{31}{20} \right)}^{2}}-\frac{1}{400} \right]\]

\[=-10\left[ {{\left( a-\frac{31}{20} \right)}^{2}}-{{\left( \frac{1}{20} \right)}^{2}} \right]=\]\[-10\left[ \left( a-\frac{31}{20}+\frac{1}{20} \right)\left( a-\frac{31}{20}-\frac{1}{20} \right) \right]\]

\[=-10\left( a-\frac{3}{2} \right)\left( a-\frac{8}{5} \right)=-(2a-3)(5a-8)\]    

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