Factorization
Category : 8th Class
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:
Multiplication |
Factorization (opposite of multiplication ) |
(1) \[2x\left( 3x-2y \right)=6{{x}^{2}}-4xy\] |
\[6{{x}^{2}}-4xy=2x\left( 3x-2y \right)\] |
(2) \[\left( 2a+3 \right)\left( 3a+2 \right)=6{{a}^{2}}+13a+6\] |
\[6{{a}^{2}}+13a+6=\left( 2a+3 \right)\left( 3a+2 \right)\] |
(3) \[\left( 15m+17n \right)\left( 15m-17n \right)=225{{m}^{2}}-289{{n}^{2}}\] |
\[225{{m}^{2}}-289{{n}^{2}}=\left( 15m+17n \right)\left( 15m-17n \right)\] |
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) \[15{{x}^{2}}{{y}^{2}}{{z}^{2}}+5x{{y}^{2}}z+5xyz=5xyz(3xyz+y+1)\]
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) \[mn{{\left( ax-2by \right)}^{2}}+m{{n}^{2}}(ax-2by)\]
\[=mn\left( ax-2by \right)\times (ax-2by+n)\]
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 1st & 3rd terms. Similarly, consider factors "ny" common between 2nd & 4th terms.
\[\Rightarrow mx\left( my+nx \right)+ny\left( nx+my \right)\]\[=\left( mx+ny \right)\times \left( nx+my \right)\]
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 \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( ax+by \right)}^{2}}\]
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) \[{{m}^{2}}{{y}^{2}}-81{{n}^{2}}\]
\[={{(my)}^{2}}-{{(9n)}^{2}}={{(my-9n)}^{2}}\]
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.
\[\therefore 6{{x}^{2}}+7x+2=6{{x}^{2}}+3x+4x+2\]
\[=3x\left( 2x+1 \right)+2\left( 2x+1 \right)=\left( 3x+2 \right)\left( 2x+1 \right).\]
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