Factorization is exactly opposite of multiplication. In multiplication we multiply two or more terms or numbers to get one term or number.
Examples for multiplication are given below:
Ex1: 3*3=9
Ex2: 2a*a=2a^{2}
The Process of writing an algebraic expression as a product of two or more expressions is called factorization.
We know that, FACTORIZATION is exactly opposite of multiplication.
Let us study factorization, by simple examples…
Ex1: Write the factors of 6.
6 = 2*3 (factors of any number must be in its simplest form)
So, 2 and 3 are the factors of 6.
Ex2: Write the factors of 10x
10x = 2*5*x
So, the factors of 10x are 2, 5 and x.
We can also take 1 as a factor of any given expression. But it does not give anything new.
Ex1: write the factors of a+8
(a+8)=1*(a+8)
This is called trivial factorization.
DIFFERENT METHODS OF FACTORIZATION – Factorization
There are many ways of factorizing a given expression. Let us study them one by one in brief.

Factorization taking common factors:
In this type of factorization we need to take common factors of a given expressions and them we need to simplify and write it again.
Example1: Write the factors of 5x^{2 }10x
Ans:
5x^{2 }10x = 5*x*x2*5*x
= 5*x*(x2)
=5x(x2)
[Note for Example1: Write the factors of 5x^{2 }10x
Step (i): write the factors of each term
5x^{2 }= 5*x*x
10x=2*5*x
Step (ii): Substitute the factors each term in the given expression.
5x^{2 }10x = 5*x*x2*5*x
Step (iii): take the common factor outside the bracket.
5x^{2 }10x = 5*x*(x2)
=5x(x20)]
Example2: factorize 4a+12b.
Ans:
4a+12b = 4*a+3*4*b
= 4*(a+3*b)
=4(a+3b)
Note for Example2: Write the factors of 4a+12b
Step (i): write the factors of each term
4a=4*a
12b=3*4*b
Step (ii): Substitute the factors each term in the given expression.
4a+12b = 4*a+3*4*b
Step (iii): take the common factor outside the bracket.
4a+12b= 4*(a+3*b)
=4(a+3b)
Example3: Factorise 3x^{2}y6xy^{2}+9xy
Ans:
3x^{2}y6xy^{2}+9xy= 3*x*x*y2*3*x*y*y+3*3*x*y
=3*x*y*(x2y+3)
=3xy(x2y+3)
Example4: Factorise a^{3}a^{2}+a.
Ans:
a^{3 }a^{2 }+a = a *a *a a *a +a
=a* (a*aa+1)
=a (a^{2}a+1)
=a (a2a+1)
Note for Example4: Write the factors of a^{3}a^{2}+a
Step (i): write the factors of each term
a^{3}=a*a*a
a^{2}=a*a
a = a
Step (ii): Substitute the factors each term in the given expression.
a^{3 }a^{2 }+a = a*a*aa*a +a
Step (iii): take the common factor outside the bracket.
a^{3 }a^{2 }+a =a*(a*aa+1)
=a (a^{2}a+1)

Factorization by grouping:
In this type of factorization we need to arrange the terms of the given expression in such a way that we can take common factors in each group. Let us study it with examples.
Example 5: Factorise axbx+ayby
Solution:
axbx+ayby= (axbx)+(ayby)
= x (ab)+y(ab)
= (ab)(x+y)
Note for Example 5: Factorise axbx+ayby
Step (i): Arrange the terms as such to group in order to take common factors
axbx+ayby = (axbx)+(ayby)
Step (ii): factorise each term
axbx+ayby=(a*xb*x)+(a*yb*y)
Step (iii): take the common factor outside the bracket.
axbx+ayby=x(ab)+y(ab)
= (ab)(x+y)
Example 6: Get the factors of y33y2+2y6xy+3x
Solution:
y^{3}3y^{2}+2y6xy+3x = (y^{3}3y^{2})+(2y6)(xy3x)
= y^{2}*(y3)+2*(y3)x*(y3)
=( y^{2}+2x)(y3)
Note for Example 5:Factorise y^{3}3y^{2}+2y6xy+3x
Step (i): Arrange the terms as such to group in order to take common factors
y^{3}3y^{2}+2y6xy+3x=(y^{3}3y^{2})+(2y6)(xy3x)
Step (ii): factorise each term
y^{3}3y^{2}+2y6xy+3x= y^{2}*(y3)+2*(y3)x*(y3)
Step (iii): take the common factor outside the bracket.
y^{3}3y^{2}+2y6xy+3x =( y^{2}+2x)(y3)

Factorization of difference of two squares
We know the formula (a + b) (ab) = (a^{2}b^{2}), for all a and b.
This leads to a nice factorization when the given expression can be written as difference of two squares.
Example 7: Factorise 36a^{2}49b^{2}.
Solution:
We know, 36a^{2}= (6a)^{2}
49b^{2}= (7b)^{2 }
36a^{2}49b^{2}= (6a)^{2}(7b)^{2}.
Thus, 36a^{2}49b^{2}=(6a+7b)(6a7b)
Note for Example 7: Factorise 36a^{2}49b^{2}.
Solution:
We know, 36a^{2}= (6a)^{2}
49b^{2}= (7b)^{2 }
36a^{2}49b^{2}= (6a)^{2}(7b)^{2}.
Now, apply the formula (a+b)(ab)=a^{2}b^{2} ,
here a=(6a)^{2} and b=(7b)^{2}
Thus we get, 36a^{2}49b^{2}= (6a)^{2}(7b)^{2}
= (6a+7b)(6a7b)
Example 8: Factorise (^{x}/_{y})^{2}(^{ 9}/_{16})^{2}
Solution:
(^{x}/_{y})^{2}(^{ 9}/_{16}) = (^{x}/_{y})^{2}(^{ 3}/_{4})^{2}
(^{x}/_{y})^{2}(^{ 9}/_{16})^{2} = [(^{x}/_{y})+(^{ 3}/_{4})][ (^{x}/_{y})(^{ 3}/_{4})]
Factorization – Exercise 2.2.2
1) Resolve into factors:
(i) x^{2}+xy
Solution:
x^{2}+xy = x(x+y) (hint: take common factor outside)
(ii) 3x^{2}6x
Solution:
3x^{2}6x = 3x(x2) (hint: take common factor outside)
(iii) (1.6)a^{2}(0.8)a
Solution:
(1.6) a^{2}(0.8) a = 16a^{2}8a (hint: multiply the given problem with 10,for easy simplification)
(1.6) a^{2}(0.8) a=8a(2a1) (hint: take common factor outside)
(iv) 510m20n
Solution:
510m20n=5(12m4n) (hint: take common factor outside)
 Factorise:
(i) a^{2}+ax+ab+bx
Solution:
a^{2}+ax+ab+bx = (a^{2}+ab) + (ax+bx) (hint: group in order to get common factor in each group)
= a(a+b) + x(a+b) (hint: take common factor outside)
= (a+x)(a+b) (hint: Simplify)
(ii) 3ac+7bc3ad7bd
Solution:
3ac+7bc3ad7bd = (3ac3ad)+(7bc7bd) (hint: group in order to get common factor in each group)
= 3a(cd)+7b(cd) (hint: take common factor outside)
= (3a+7b)(cd) (hint: simplify)
(iii) 3xy6zy3xt+6zt
Solution:
3xy6zy3xt+6zt = (3xy3xt)+(6zt6zy) (hint: group in order to get common factor in each group)
= 3x(yt)+6z(ty) (hint: take common factor outside)
= (3x+6z)(yt) (hint: simplify)
(iv) y^{3}3y^{2}+2y6xy+3x
Solution:
y^{3}3y^{2}+2y6xy+3x = ( y^{3}xy+2y)+(3y^{2}6+3x)
= y(y^{2}x+2)3(y^{2}x+2)
= (y3)( y^{2}x+2)
 Factorise
(i) 4a^{2}25
Solution:
4a^{2}25 = (2a)^{2}(5)^{2 }(hint: write the given problem as perfect squares)
Apply the formula a^{2}b^{2 }= (a+b)(ab)
Here a = (2a) and b = 5
Substitute the value of a and b in the formula a^{2}b^{2 }= (a+b)(ab)
Now, 4a^{2}25 = (2a+5)(2a5)
(ii) x^{2}9/16
Solution:
x^{2}9/16= x^{2}(
=x^{2}–
Apply the formula a^{2}b^{2 }= (a+b)(ab)
Here a= x^{ }and b=(
Substitute the value of a and b in the formula a^{2}b^{2 }= (a+b)(ab)
Now, x^{2}9/16 = (x+(x
(iii) x^{4}y^{4}
Solution:
x^{4}y^{4} = ( x^{2})^{2}( y^{2})^{2}
Apply the formula a^{2}b^{2 }= (a+b)(ab)
Here a = x^{2} and b = ( y^{2})
Substitute the value of a and b in the formula a^{2}b^{2 }= (a+b)(ab)
x^{4}y^{4} = (x^{2}– y^{2} )( x^{2}+ y^{2} )
(iv) (7)^{2}(2)^{2}
Solution:
(7)^{2}(2)^{2}
Apply the formula a^{2}b^{2 }= (a+b)(ab)
Here a = (7) and b = (2)
Substitute the value of a and b in the formula a^{2}b^{2 }= (a+b)(ab)
(7)^{2}(2)^{2 }= (7+2
(v) (0.7)^{2}(0.3)^{2 }
Solution:
(0.7)^{2}(0.3)^{2 }
Apply the formula a^{2}b^{2 }= (a+b)(ab)
Here a = 0.7 and b = 0.3
Substitute the value of a and b in the formula a^{2}b^{2 }= (a+b)(ab)
(0.7)^{2}(0.3)^{2 } = (0.7+0.3)(0.70.3)
(vi) (5a2b)^{2}(2ab)^{2}
Solution:
Apply the formula a^{2}b^{2 }= (a+b)(ab)
Here a = (5a2b) and b = (2ab)
Substitute the value of a and b in the formula a^{2}b^{2 }= (a+b)(ab)
(5a2b)^{2}(2ab)^{2 }= ((5a2b)+( 2ab))(5a2b)( 2ab))
Factorization of trinomials – Factorization:
We have learnt earlier how to multiply two binomials of the form (x+a) and (x+b):
(x+a)(x+b) = x^{2}+(a+b)x+ab
We can also proceed in the reverse direction. Given the trinomial of the form x^{2}+(a+b)x+ab, we can factorise this to get x^{2}+(a+b)x+ab = (x+a)(x+b). But generally, the trinomial is not given in this form. You may be given in the form x^{2}+mx+n, where m, n are some numbers. Let us study them here,
The sum and product of two numbers are positive if and only if both the numbers are positive.
This says that if a+b and ab are positive then so are a and b. The converse is also true. Thus , 6 and 5 are positive; 5=3+2 and 6=3*2; both 3 and 2 are positive.
The sum of two numbers is negative and their product positive if and only if both the numbers are negative.
Thus a+b negative and ab positive if and only if a and b are negative. If we are given numbers 21 and 10, we see that 10= (7) + (3) and 21= (7) * ( 3).
We say 7 is the absolute value of both 7 and 7. Thus given an integer a, we define its absolute value by a= a if a>0; a= a if a<0; and a=0 if a=0. Observe 8<6 but 8=8>6=6.
The sum of two numbers is positive and their product is negative if and only if the numbers is positive and the other negative, and the positive number has larger absolute value than the negative.
This means a+b is positive and ab negative only if one of a,b is positive and other negative; and if a is positive and b is negative , then a<b. For example, we see that if a+b = 7 and ab = 18,then a = 9 and b = 2 or a = 2 and b = 9.
The sum of two numbers is negative and their product negative if and only if one of the numbers is positive and the other negative, and the positive has smaller absolute value than the negative number.
Thus a+b is negative and ab negative if and only if one a,b is positive and other is negative; if a is positive and b is negative, then a>b. For example, if a+b = 12 and ab = 28, we can write a = 2 and b = 14 or a = 14 and b = 2.
Here we have not mentioned anything about the nature of numbers. They can be integers, rational numbers or even real numbers.
[Note: factorization of trinomials will not given in this form , it may be given the form x^{2}+px+q, where p and q are some numbers. We must have to bring the given problem to the form p= (a+b) and q = ab, to factorise the trinomials.]
Example 10: factorise 6x^{2}+11x+3
Solution:
6x^{2}+11x+3
We have, 6*3 = 18 [We have to multiply 6 and 3 which is equal to 18]
Choose any two numbers such that, sum of those numbers must be 11 and product of the same numbers must be 18.
So, we have 9 and 2, which satisfies the condition for the given problem 6x^{2}+11x+3.
i.e.,
9+2 = 11 and 9*2 = 19.
Therefore,
6x^{2}+11x+3 = 6x^{2}+(9x+2x)+3 [hint: 9x+2x = 11x = p ; 9x*2x = 18x^{2 }= q]
= ( 6x^{2}+9x)+(2x+3)
= 3x (2x+3) +1(2x+3)
= (2x+3) (3x+1)
Example11: factorise x^{2}9x+20
Solution:
x^{2}9x+20
we have, 1*20 = 20.
Choose any two numbers such that, sum of those numbers must be 9 and product of the same numbers must be 20.
So, we have 4 and 5,which satisfies the condition for the given problem x^{2}9x+20.
i.e.,
(4) + (5) = (9) and (4)*(5) = (+20).
Therefore,
x^{2}9x+20 = x^{2}4x5x+20
= (x^{2}4x)(5x20)
= x(x4)5(x4)
= (x4)(x5)
Factorising a sqaure trinomial – Factorization:
Any algebraic expression which can be written in the form of a^{2} – 2ab + b^{2} = (a – b)^{2} is called a square trinomial.
For example, x^{2}+2x+1 is a square trinomial. We can have immediate factorisation for such a trinomial using a^{2}+2ab+b^{2}=(a+b)^{2} or a^{2}2ab+b^{2}=(ab)^{2} .
Example12: factorise 4x^{2} – 12xy + 9y^{2}
Solution:
The given problem 4x^{2} – 12xy + 9y^{2} is in the form a^{2} – 2ab + b^{2}
Here, a=2x and b=3y.
Substitute the value of a and b in the formula to factorise.
4x^{2} – 12xy + 9y^{2} = (2x)^{2} – 2 (2x)(3y) + (3y)^{2}
We know, a^{2} – 2ab + b^{2} = (a – b)^{2}
Therefore, 4x^{2} – 12xy + 9y^{2} = (2x – 3y)^{2}
Example 13: is x^{2} – 6xy + 36y^{2} a square trinomial?
Solution:
We observe that x^{2} – 6xy + 36y^{2} = x^{2} – x(6y) + (6y)^{2}
To be a square trinomial, given equation should be in the form of the formula a^{2} + 2ab + b^{2} = (a + b)^{2} but we have,
x^{2} – 6xy + 36y^{2} = x^{2} – x(6y) + (6y)^{2} which is not in standard form.
Thus, x^{2} – 6xy + 36y^{2} is not a square trinomial.
Factorization – Exercise 2.2.3
 In the following, you are given pq and the sum p+q.
Determine p and q.
(i) pq = 18 and p+q = 11
Solution:
it is given that pq = 18 and p+q = 11.
So, p = 2 and q = 9, which satisfies the condition.
(ii) pq = 32 and p+q = 12
Solution:
It is given that pq = 32 and p+q = 12.
So, p = 8 and q = 4, which satisfies the condition.
(iii) pq = 24 and p+q = 2
Solution:
It is given that pq = 24 and p+q = 2.
So, p = 6 and q = 4, which satisfies the condition.
(iv) pq = 12 and p+q = 11
Solution:
it is given that pq = 12 and p+q = 11.
So, p = 12 and q = 1, which satisfies the condition.
(v) pq = 6 and p+q = 5
Solution:
It is given that pq = 6 and p+q = 5.
So, p = 6 and q = 1, which satisfies the condition.
(vi) pq = 44 and p+q = 7
Solution:
It is given that pq = 44 and p+q = 7.
So, p = 11 and q = 4, which satisfies the condition.
 Factorise
(i) x^{2}+6x+8
Solution:
x^{2}+6x+8 = x^{2}+(4x+2x)+8
= (x^{2}+4x) + (2x+8)
= x(x+4) +2 (x+4)
= (x+2)(x+4)
(ii) 4x^{2}+4x+1
Solution:
4x^{2}+4x+1 = 4x^{2}+ (2x+2x) + 1
= (4x^{2}+2x) + (2x+1)
= 2x (2x+1)+1(2x+1)
= (2x+1) (2x+1)
(iii) a^{2}+5a+6
Solution:
a^{2}+5a+6 = a^{2}+ (2a+3a)+6
= (a^{2}+2a)+(3a+6)
= a(a+2)+3(a+2)
= (a+2)(a+3)
(iv) a^{2}5a+6
Solution:
a^{2}5a+6 = a^{2}(3a2a)+6
= (a^{2}3a)(2a6)
= a (a3)2(a3)
= (a3)(a2)
(v)a^{2}3a40
Solution:
a^{2}3a40 = a^{2}8a+5a40
= (a^{2}8a)+(5a40)
= a(a8)+5(a8)
= (a8)(a+5)
(vi) x^{2}x72
Solution:
x^{2}x72 = x^{2}8x+7x72
= (x^{2}8x)+(7x72)
= x(x8)+7(x8)
= (x8)(x+7)
 Factorise
(i) x^{2}+14x+49
Solution:
Use the formula a^{2} + 2ab + b^{2} = (a + b)^{2}
x^{2}+14x+49 = x^{2} +2(x)(7)+7^{2}
Here a = x and b = 7
Substitute the values of a and b in the formula,
x^{2}+14x+49 = ( x+7)^{2}^{ }
(ii) 4x^{2}+4x+1
Solution:
Use the formula a^{2} + 2ab + b^{2} = (a + b)^{2}
4x^{2}+4x+1 = 4x^{2}+2(2x)(1)+1
Here a = 2x and b = 1
Substitute the values of a and b in the formula,
4x^{2}+4x+1 = ( 2x+1)^{2}
(iii) a^{2}10a+25
Solution:
Use the formula a^{2} – 2ab + b^{2} = (a – b)^{2}
a^{2}10a+25 = a^{2}2(a)(5)+ 5^{2}
Here a = a and b = 5
Substitute the values of a and b in the formula,
x^{2}+14x+49 = ( a+5)^{2}
(iv) 2x^{2}24x+72
Solution:
Use the formula a^{2} – 2ab + b^{2} = (a – b)^{2}
2x^{2}24x+72 = 2(x^{2}12x+36) = 2(x^{2}2(x)(6)+36)
Here a = x and b = 6
Substitute the values of a and b in the formula,
2x^{2}24x+72=2(x^{2}6x+36) = 2(x6)^{2}
(v) p^{2}24p+144
Solution:
Use the formula a^{2} – 2ab + b^{2} = (a – b)^{2}
p^{2}24p+144 = p^{2}2(p)(12)+ 12^{2}
Here a = p and b = 12
Substitute the values of a and b in the formula,
p^{2}24x+144 = ( p12)^{2}
(vi) x^{3}12x^{2}+36x
Solution:
Use the formula a^{2} – 2ab + b^{2} = (a – b)^{2}
x^{3}12x^{2}+36x = x(x^{2}12x+36) = x(x^{2}6x6x+36)
x^{3}12x^{2}+36x = x(x^{2}—2(6)(x)+6^{2})
Here a = 6 and b = 6
Substitute the values of a and b in the formula,
x^{3}12x^{2}+36x = x(x6)^{2}
Points to remember – Factorization:
 The Process of writing an algebraic expression as a product of two or more expressions is called factorization.
 FACTORIZATION is exactly opposite of multiplication.
 We can also take 1 as a factor of any given expression. But it does not give anything new this is called trivial factorization
 Methods of factorization are
 factorization by taking common factors
 factorization by grouping
 factorization of difference of two squares
Formula used in Factorization:
a^{2} – 2ab + b^{2} = (a – b)^{2}
a^{2} + 2ab + b^{2} = (a + b)^{2}
(a + b)(a – b) = (a – b)^{2}
x^{2 }+ px + q, you must find numbers a and b such that a.b = q and a+b = p. Then x^{2 }+ px + q = (x+a) (x+b).
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