Sachin

Factorize by grouping the terms consists of two or more product terms as resultant value. To divide the expression into product terms, we need to identify the greatest common factor which divides all the remaining terms in the expression. Step by step process on how to do Factorization by Grouping the Terms are clearly given in this article.

Procedure to Factorize by Grouping the Terms

Refer to the below-mentioned step by step process and learn the Factoring by Grouping Terms. They are along the lines

1. Note down the given expression.
2. Group the first two terms and last two terms.
3. Factor out the greatest common factor from each group.
4. Finally, you will get two or more product terms as result.

Solved Examples on Factorization by Grouping Terms

1. Factorize grouping the following expressions?

(i) 18x³y³ – 27x²y³ + 36x³y²

Solution:
The given expression is 18x³y³ – 27x²y³ + 36x³y²
Factor out the greatest common factor from the above expression.
That is, 9x²y²(2xy – 3y + 4x).

Therefore, the final solution for the expression 18x³y³ – 27x²y³ + 36x³y² is 9x²y²(2xy – 3y + 4x).

(ii) 12a²b³ – 21a³b²

Solution:
The given expression is 12a²b³ – 21a³b²
Factor out the greatest common factor from the above expression.
3a²b²(4b – 7a).

Finally, the solution for the expression 12a²b³ – 21a³b² is 3a²b²(4b – 7a).

(iii) a³ – a² + a – 1.

Solution:
The given expression is a³ – a² + a – 1
Group the first two terms and last two terms.
Here, first two terms are a³ – a² and the last two terms are a – 1.
So, (a³ – a²) + (a – 1).
Now, factor out the greatest common factor from each group.
That is, a²(a – 1) + (a – 1).
(a – 1) (a² + 1).

Therefore, solution for the expression a³ – a² + a – 1 is (a – 1) (a² + 1).

(iv) prs + qurs – pt – qut

Solution:
The given expression is prs + qurs – pt – qut.
Group the first two terms and last two terms.
Here, the first two terms are prs + qurs and last two terms are – pt – qut.
Then, (prs + qurs) – (pt + qut).
Now, factor out the greatest common factor from the above two groups.
That is, rs(p + qu) – t(p + qu).
(p + qu) (rs – t).

Therefore, the solution for the expression prs + qurs – pt – qut is (p + qu) (rs – t).

(v) a² – 3a – ab + 3b
Solution:

The given expression is
a² – 3a – ab + 3b.
Group the first two terms and last two terms.
Here, the first two terms are a²-3a and the last two terms are ab + 3b.
Then, (a² – 3a) – (ab – 3b).
Now, factor out the greatest common factor from the above two groups.
That is a(a – 3) – b(a – 3).
(a – 3) (a – b).

Therefore, solution for the expression is a² – 3a – ab + 3b is (a – 3) (a – b).

2. How to factorize by grouping the following expressions?
(i) 2q^4 – q³ + 4q – 2
Solution:
The given expression is
2q^4 – q³ + 4q – 2.
Group the first two terms and last two terms.
Here, the first two terms are 2q^4 – q³ and the last two terms are 4q – 2.
Then, (2q^4 – q^3) + (4q-2).
Now, factor out the greatest common factor from the above two groups.
That is, q^3(2q – 1) +2(2q – 1).
(2q – 1) (q^3 + 2).

Therefore, solution for the expression is 2q4 – q3 + 4q – 2 is (2q-1) (q^3+2).

(ii) ac + bc – ad – bd.

Solution:
The given expression is ac + bc – ad – bd.
Group the first two terms and last two terms.
Here, the first two terms are ac+bc and the last two terms are – ad-bd.
Then, (ac + bc) – (ad + bd).
Now, factor out the greatest common factor from the above two groups.
That is, c(a+b) -d(a+b).
(a+b) (c-d).

Therefore, solution for the expression is ac + bc – ad – bd is (a+b) (c-d).

(iii) pa – pb – qa –qb.

Solution:
The given expression is pa – pb – qa–qb.
Group the first two terms and last two terms.
Here, the first two terms are pa – pb, and the last two terms are qa – qb.
Then, (pa – pb) – (qa -qb).
Now, factor out the greatest common factor from the above two groups.
That is, p(a – b) – q(a – b).
(p – q) (a – b).

Therefore, the solution for the expression is pa – pb – qa – qb is (p – q) (a – b).

3. How to factorize by grouping the algebraic expressions?
(i)x^2z^2 + xzw + xyz + yw

Solution:
The given expression is
x^2z^2 + xzw + xyz + yw.
Group the first two terms and last two terms.
Here, the first two terms are x^2z^2 + xzw and the last two terms are xyz + yw.
Then, (x^2z^2 + xzw) + (xyz + yw).
Now, factor out the greatest common factor from the above two groups.
That is, xz(xz + w) + y(xz + w).
(xz + y) (xz + w).

Therefore, solution for the expression is x^2z^2 + xzw + xyz + yw is (xz + y) (xz + w).

(ii) 5x + xy + 5y + y^2

Solution:
The given expression is 5x + xy + 5y + y^2
Group the first two terms and last two terms.
Here, the first two terms are 5x + xy, and the last two terms are 5y + y^2.
Then, (5x + xy) + (5y + y^2).
Now, factor out the greatest common factor from the above two groups.
That is, x(5 + y) + y(5 + y).
(5 + y) (x + y).

Therefore, solution for the expression is 5x + xy + 5y + y^2 is (5 + y) (x + y).

(iii) xy – yz – xz + z^2

Solution:
The given expression is xy – yz – xz + z^2
Group the first two terms and last two terms.
Here, the first two terms are xy – yz and the last two terms are – xz + z^2.
Then, (xy – yz) – (xz – z^2).
Now, factor out the greatest common factor from the above two groups.
That is, y(x – z) – z(x – z).
(x – z) (y – z).

Therefore, solution for the expression xy – yz – xz + z^2 is (x – z) (y – z).

4. Factorize the expressions
(i) a^4 + a^3 + 2a + 2

Solution:
The given expression is a^4 + a^3 + 2a + 2.
Group the first two terms and last two terms.
Here, the first two terms are a^4 + a^3 and the last two terms are 2a + 2.
Then, (a^4 + a^3) + (2a + 2).
Now, factor out the greatest common factor from the above two groups.
That is, a^3(a + 1) + 2(a + 1).
(a + 1) (a^3 + 2).

Therefore, solution for the expression a^4 + a^3 + 2a + 2 is (a + 1) (a^3 + 2).

(ii) a²b² + d²b² – cd² – ca²

Solution:
The given expression is a²b² + d²b² – cd² – ca²
Group the first two terms and last two terms.
Here, the first two terms are a²b² + d²b² and the last two terms are – cd² – ca².
Then, (a^2b^2 + d^2b^2) – (cd^2 + ca^2).
Now, factor out the greatest common factor from the above two groups.
That is, b^2(a^2 + d^2) – c(d^2 + a^2).
(a^2 + d^2) (b^2 – c).

Therefore, solution for the expression a2b2 + d2b2 – cd2 – ca2 is(a^2 + d^2) (b^2 – c).

5. Factorize by grouping the terms
(x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y

Solution:
Thegiven expression is (x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y.
Group the first two terms and last two terms.
here, first two terms are (x^2 + 3x)^2 – 2(x^2 + 3x) and the last two terms are –y(x^2 + 3x) + 2y.
Then, [(x^2 + 3x)^2 – 2(x^2 + 3x)] –[y(x^2 + 3x) – 2y].
That is, (x^2 + 3x)(x^2 + 3x – 2) – y(x^2 + 3x – 2).
(x^2 + 3x – 2)(x^2 + 3x – y).

Therefore, solution for the expression (x2 + 3x)2 – 2(x2 + 3x) – y(x2 + 3x) + 2y is (x^2 + 3x – 2)(x^2 + 3x – y).

Are you looking for the different problems on Factorization by Grouping? Then, you are in the right place. We have given all types of factorization problems on our website. Students can learn to Factorize by Grouping the Terms in these articles. While solving Factorization by grouping problems, students need to group the terms with common factors before factoring.

How to Factor by Grouping?

Have a look at the Factoring by Grouping Steps and learn how to solve related problems easily. Follow the guidelines provided and perform factorization by grouping method. They are as under

• Note down the given expression, group the first two terms and last two terms.
• Find out the greatest common factor(GCF) from the first term and second term.
• Now, find the common factor from the above two groups.
• Finally factor out the terms in terms of product.

Factorization by Grouping Examples

1. Factor grouping the expressions?
1 + x + xy + x²y.

Solution: Given Expression is 1 + x + xy + x²y.
Group the first two terms and last two terms.
First two terms are 1 + x and the last two terms are xy + x²y.
(1+ x) + (xy + x²y).
Find out the common factor from the above two groups.
(1+x) + xy(1+x).
Factor out the terms in terms of product.
(1+x) (1+xy).

By factor grouping the expression 1 + x + xy + x²y., we will get (1+x) (1+xy).

2. How to factor by grouping the following algebraic expressions?

(i) x² – xy + xz – zy.

Solution:
Given Expression is x² – xy + xz – zy
Group the first two terms and last two terms.
First two terms are x² – xy and the last two terms are xz – zy.
(x² – xy) + (xz – zy)
Find out the common factor from the above two groups.
x(x – y) + z(x – y)
Factor out the terms in terms of product.
(x + z) (x – y)

Factor by grouping the expression x² – xy + xz – zy, we will get the result as (x+z) (x-y).

(ii) x² + 3x + xy + 3y.

Solution:
Given Expression is x² + 3x + xy + 3y.
Group the first two terms and last two terms.
First two terms are x² + 3x and the last two terms are xy + 3y.
(x² + 3x) + (xy + 3y)
Find out the common factor from the above two groups.
x(x + 3) + y(x + 3)
Factor out the terms in terms of product.
(x+y) (x+3).

Factor by grouping the expression x² + 3x + xy + 3y, we will get solution as (x+y) (x+3).

3. Factorize the algebraic expressions.

(i) 2a + ba + 2b + b²

Solution:
Given Expression is 2a + ba + 2b + b²
Group the first two terms and last two terms.
First two terms are 2a + ba and the last two terms are 2b + b².
(2a + ba)+ (2b + b²).
Find out the common factor from the above two groups.
a(2 + b) + b(2 + b).
Factor out the terms in terms of product.
(a + b) (2 + b).

By factorizing the expression 2a + ba + 2b + b², we will get (a + b) (2 + b).

(ii) b² – yb + 5b– 5y.

Solution:
Given Expression is b² – yb + 5b– 5y.
Group the first two terms and last two terms.
First, two terms are b² – yb and the last two terms are 5b – 5y.
(b² – yb)+ (5b – 5y).
Find out the common factor from the above two groups.
b(b – y) + 5(b – y).
Factor out the terms in terms of product.
(5 + b) (b – y).

By factorizing the expression b² – yb + 5b– 5y, we will get (5 + b) (b – y).

(iii) pq – rq – ps + rs.

Solution:
Given Expression is pq – rq – ps + rs.
Group the first two terms and last two terms.
First two terms are pq – rq and the last two terms are – ps + rs.
(pq – rq) – (ps – rs).
Find out the common factor from the above two groups.
q(p – r) -s(p – r).
Factor out the terms in terms of product.
(p – r) (q – s).

By factorizing the expression pq – rq – ps + rs, we will get (p – r) (q – s).

(iv) ab – 2ac – db + 2dc.

Solution:
The given expression is ab – 2ac – db + 2dc.
Group the first two terms and last two terms.
Fist two terms are ab – 2ac and the last two terms are – db + 2dc.
(ab – 2ac) – (db – 2dc).
Find out the common factor from the above two groups.
a(b – 2c) -d(b – 2c).
Factor out the terms in terms of product.
(b – 2c) (a – d).

By factorizing the expression ab – 2ac – db + 2dc., we will get (b – 2c) (a – d).

(v) pq² – 3rqs – pqs + 3rs²

Solution:
The given expression is pq² – 3rqs – pqs + 3rs²
Group the first two terms and last two terms.
First two terms are pq^2 – 3rqs and the last two terms are –pqs + 3rs².
(pq² – 3rqs) – (pqs – 3rs²).
Find out the common factor from the above two groups.
q(pq – 3rs) -s(pq – 3rs).
Factor out the terms in terms of product.
(pq – 3rs) (q – s).

By factorizing the expression pq² – 3rqs – pqs + 3rs², we will get (pq – 3rs) (q – s).

4. Factor each of the following expressions by grouping

(i) a² – 3a – ab + 3b.

Solution:
Given expression is a² – 3a – ab + 3b
Group the first two terms and last two terms.
First two terms are a² – 3a and the last two terms are – ab + 3b.
(a² – 3a) – (ab – 3b).
Find out the common factor from the above two groups.
a(a – 3) – b(a – 3).
Factor out the terms in terms of product.
(a – 3) (a – b).

By factorizing the expression a² – 3a – ab + 3b, we will get (a – 3) (a – b).

(ii) pq² + rq² + 2p + 2r.

Solution:
Given expression is pq² + rq² + 2p + 2r
Group the first two terms and last two terms.
First two terms are pq² + rq² and the last two terms are 2p + 2r.
(pq² + rq²) + (2p + 2r).
Find out the common factor from the above two groups.
q²(p +r) + 2(p+r).
Factor out the terms in terms of product.
(q²+ 2) (p + r).

By factorizing the expression pq² + rq² + 2p + 2r, we will get (q²+ 2) (p + r).

(iii) 2pq² + 3pqr – 2sqr – 3sr²

Solution:
Given expression is 2pq² + 3pqr – 2sqr – 3sr²
Group the first two terms and last two terms.
First two terms are 2pq² + 3pqr and the last two terms are – 2sqr – 3sr².
(2pq² + 3pqr) – ( 2sqr + 3sr²).
Find out the common factor from the above two groups.
pq(2q + 3r) – sr(2q + 3r).
Factor out the terms in terms of product.
(2q + 3r) (pq – sr).

By factorizing the expression 2pq² + 3pqr – 2sqr – 3sr², we will get (2q + 3r) (pq – sr).

(iv) par² + qars – pnrs – qns²

Solution:
The given expression is par² + qars – pnrs – qns²
Group the first two terms and last two terms.
First two terms are par² + qars and the last two terms are – pnrs – qns².
(par² + qars) – (pnrs – qns²).
Find out the common factor from the above two groups.
ar(pr + qs) – ns(pr + qs).
Factor out the terms in terms of product.
(pr + qs) (ar – ns).

By factorizing the expression par² + qars – pnrs – qns², we will get (pr + qs) (ar – ns).

5. Factorize
(i) (a + b) (2a + 5) – (a + b) (a + 3).

Solution:
The given expression is (a + b) (2a + 5) – (a + b) (a + 3).
Find out the common factor from the above expression.
(a + b) [(2a + 5) – (a + 3)].
Expand the terms.
(a + b)(2a + 5 – a – 3).
Simplify the second term.
(a + b) (a + 2).

By factorizing the expression (a + b) (2a + 5) – (a + b) (a + 3), we will get (a + b)(a + 2).

(ii) 6xy – y² + 12xz – 2yz.

Solution:
The given expression is 6xy – y² + 12xz – 2yz.
Group the first two terms and last two terms.
First two terms are 6xy – y² and the last two terms are 12xz – 2yz.
(6xy – y²) + (12xz – 2yz).
Find out the common factor from the above two groups.
y(6x – y)+ 2z(6x – y).
Factor out the terms in terms of product.
(6x – y) (y + 2z).

By factorizing the expression 6xy – y² + 12xz – 2yz, we will get (6x – y) (y + 2z).

Students can find the factors for an algebraic expression when one of its factors is binomial. Find given algebraic expression Factorization when Binomial is Common Factor. All related problems are included in this article along with a clear explanation. Therefore, students can practice every problem available in this article and improve their knowledge. The process of solving the factorization problem is very simple if you follow the procedure we explained below. Go through the complete article and learn different methods to solve factorization problems.

Factorization of Algebraic Expressions when Binomial is Common

Follow the simple and easy guidelines on Factorization of Algebraic Expressions When Binomial is Common. They are as follows

Step 1: In the first step, find the common binomial factor.
Step 2: Note down the given expression as the product of this binomial and the quotient obtained on dividing the given expression by this binomial.

Solved Examples on Factorization When Binomial is a Common

1. Factorize the algebraic expressions.

(i) 7b(3x – 4y) + 3a(3x – 4y)

Solution:
Given expression is 7b(3x – 4y) + 3a(3x – 4y)
In the given expression, the binomial factor is (3x – 4y) as it is common in both terms.
Take the (3x – 4y) common and multiply it with the remained terms.
(3x – 4y) (7b + 3a)

The final answer is (3x – 4y) (7b + 3a)

(ii) 12(9a + 6b)² – 4(9a + 6b)

Solution:
Given expression is 12(9a + 6b)² – 4(9a + 6b)
12 (9a + 6b) (9a + 6b) -4(9a + 6b)
In the given expression, the binomial factor is (9a + 6b) as it is common in both terms.
Take the (9a + 6b) common and multiply it with the remained terms.
(9a + 6b)(12(9a + 6b) – 4)
(9a + 6b)(108a + 72b – 48)
The final answer is (9a + 6b)(108a + 72b – 48)

2. Factorize the expression 10r(m – 2n) – 8m + 16n

Solution:
Given expression is 10r(m – 2n) – 8m + 16n
Lets take -8m  + 16n from the above equation.
Take -8 common from -8m + 16n
-8(m – 2n)
Place -8(m – 2n) in 10r(2m – 4n) – 8m + 16n equation.
10r(m – 2n) -8(m – 2n)
In the above expression, the binomial factor is (m – 2n) as it is common in both terms.
Take the (m – 2n) common and multiply it with the remained terms.
(m – 2n)(10r – 8)

The final answer is (m – 2n)(10r – 8)

3. Factorize (a – 4b)^2 – 7a + 28b

Solution:
(a – 4b)(a – 4b) – 7a + 28b
Given expression is (a – 4b)(a – 4b) – 7a + 28b
Lets take – 7a + 28b from the above equation.
Take -7 common from – 7a + 28b
-7(a – 4b)
Place -7(a – 4b) in (a – 4b)(a – 4b) – 7a + 28b equation.
(a – 4b)(a – 4b) – 7(a – 4b)
In the above expression, the binomial factor is (a – 4b) as it is common in both terms.
Take the (a – 4b) common and multiply it with the remained terms.
(a – 4b)(a – 4b – 7)

The final answer is (a – 4b)(a – 4b – 7)

All solved examples of Factorization problems when Binomial is a common factor are included in this article. Check how to find the binomial factor and practice all problems given here. Compare the terms in the given expression and factor out the greatest common factor term from the expression. Based on the greatest common factor, we will get the solution for the expression that is in terms of the product of two or more terms.

How to Factorize taking out the Binomial Common Factor?

Go through the below step by step procedure and do Factorization when a Binomial is a Common Factor. They are along the lines

1. Note down the given expression
2. Note down the first term and second term from the expression
3. Compare the two terms and observe the greatest common factor
4. Now, factor out the greatest common factor from the expression
5. Finally, we will get the result in the form of the product of two or more terms.

Solved Binomial Factor Examples

1. Factorize the expression (6x + 1)² – 5(6x + 1)?

Solution:
The given expression is (6x + 1)² – 5(6x + 1)
Here, the first term is (6x + 1)² and second term is – 5(6x + 1)
By comparing the above two terms, we can observe the greatest common factor and that is (6x + 1)
Now, factor out the greatest common factor from the expression
That is, (6x + 1) [(6x + 1) – 5]
(6x + 1)(6x – 4)

Therefore, the resultant value for the expression (6x + 1)² – 5(6x + 1) is (6x + 1)(6x – 4)

2. Factorize the algebraic expression 4x(y – z) + 3(y – z)?

Solution:
The given expression is 4x(y – z) + 3(y – z)
Here, the first term is 4x(y – z) and the second term is 3(y – z)
By comparing the above two terms, we can observe the greatest common factor and that is (y – z)
Now, factor out the greatest common factor from the expression
That is, (y – z)(4x + 3)

Therefore, the resultant value for the expression 4x(y – z) + 3(y – z) is (y – z)(4x + 3)

3. Factorize the expression (3x – 4y) (p – q) + (3x – 2y) (p – q)?

Solution:
The given expression is (3x – 4y) (p – q) + (3x – 2y) (p – q)
Here, the first term is (3x – 4y) (p – q) and the second term is (3x – 2y) (p – q)
By comparing the above two terms, we can observe the greatest common factor and that is (p – q)
Now, factor out the greatest common factor from the expression
That is, (p – q)[(3x – 4y) + (3x – 2y)]
= (p – q)(6x – 6y)

Therefore, the resultant value for the expression (3x – 4y) (p – q) + (3x – 2y) (p – q) is (p – q)(6x – 6y)

4. Factorize the expression 6(2x + y)² +8(2x + y)?

Solution:
The given expression is 6(2x + y)² +8(2x + y)
Here, the first term is 6(2x + y)² and the second term is 8(2x + y)
By comparing the above two terms, we can observe the greatest common factor and that is (2x + y)
Now, factor out the greatest common factor from the expression
That is, (2x + y)[6(2x + y) + 8]
(2x + y)(12x + 6y + 8)

Therefore, the resultant value for the expression 6(2x + y)² +8(2x + y) is (2x + y)(12x + 6y + 8)

While finding the factorization when monomial is a common factor, an algebraic expression can be expressed as the sum or the difference of monomials. Look at the different problems that have an algebraic expression with Monomial as a common factor. Also, find all Factorization concepts on our website with a detailed explanation.

How to Find Factorization of an Expression?

Below is the step by step guidelines on how to factorize an expression. They are along the lines

Step 1: Note Down the algebraic expression.
Step 2: Find the different terms of given algebraic expressions.
Step 3: Find the HCF of all the separate terms of the given algebraic expression.
Step 4: Express each term of the algebraic expression as the product of H.C.F and divide each term of the given expression by the HCF.
Step 5: Now use the distributive property of multiplication over addition or subtraction to express the algebraic expression as the product of H.C.F and the quotient of the expression divided by the H.C.F.
Step 6: In the final step, keep the H.C.F. outside the bracket and the quotients obtained within the bracket.

Solved Examples of Factorization when Monomial is Common

1. Factorize each of the following

(i) 4b + 20

Solution:
The given expression is 4b + 20
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4 and 20 is 4.
HCF of literal coefficients:
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is nothing.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4b + 20 is 4.
Multiply and divide each term of the given expression 4b + 20 with 4.
4(4b/4 + 20/4) = 4(b + 5)

The final answer is 4(b + 5)

(ii) 3m² + 2m

Solution:
The given expression is 3m² + 2m
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3 and 2 is nothing.
HCF of literal coefficients:
The lowest power of m is 1.
Therefore, the HCF of literal coefficients is m.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3m² + 2m is m.
Multiply and divide each term of the given expression 3m² + 2m with m.
m(3m²/m + 2m/m) = m(3m + 2)

The final answer is m(3m + 2)

(iii) 6a²b – 3ab²

Solution:
The given expression is 6a²b – 3ab²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 6 and 3 is 3.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
Therefore, the HCF of literal coefficients is ab.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 6a²b – 3ab² is 3ab.
Multiply and divide each term of the given expression 3m² + 2m with m.
3ab(6a²b/3ab – 3ab²/3ab) = 3ab(2a – b)

The final answer is 3ab(2a – b)

(iv) 9cd – 6bc

Solution:
The given expression is 9cd – 6bc
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 9 and 6 is 3.
HCF of literal coefficients:
The lowest power of b is 0.
The lowest power of c is 1.
The lowest power of d is 0.
Therefore, the HCF of literal coefficients is c.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 9cd – 6bc is 3c.
Multiply and divide each term of the given expression 9cd – 6bc with 3c.
3c(9cd/3c – 6bc/3c) = 3c(3d – 2b)

The final answer is 3c(3d – 2b)

2. Factorize 18x²y²z + 12xyz.

Solution:
The given expression is 18x²y²z + 12xyz
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 18 and 12 is 3.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
The lowest power of z is 1.
Therefore, the HCF of literal coefficients is xyz.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 18x²y²z + 12xyz is 3xyz.
Multiply and divide each term of the given expression 18x²y²z + 12xyz with 3xyz.
3xyz(18x²y²z/3xyz + 12xyz/3xyz) = 3xyz(6xy + 4)

The final answer is 3xyz(6xy + 4)

3. Factorize the expression: 8m³ – 32m²n

Solution:
The given expression is 8m³ – 32m²n
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 8 and 32 is 8.
HCF of literal coefficients:
The lowest power of m is 2.
The lowest power of n is 0.
Therefore, the HCF of literal coefficients is m².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 8m³ – 32m²n is 8m².
Multiply and divide each term of the given expression 8m³ – 32m²n with 8m².
8m²(8m³/8m² – 32m²n/8m²) = 8m²(m – 4n)

The final answer is 8m²(m – 4n)

Every algebraic expression has factors as 1 and itself. If the Monomial is a Common Factor, then the product of two or more numbers or variables, then it will have factors of 1 and itself. Find different problems on the Factorization of algebraic expressions and know the process to solve each problem in an easy way. We have included tips and tricks to solve Factorization Problems that help students to learn the Factorization concept easily.

How to Find Factorization of an Expression when a Monomial is a Common Factor?

Follow the below process to find the Factorization of given problems.

(i) Write an algebraic expression.
(ii) Find the H.C.F. of all the individual terms of the expression.
(iii) Divide each individual term of the expression by the H.C.F.
(iv) Keep the H.C.F. outside the bracket and the quotients obtained within the bracket.

The greatest common factor (GCF) of two or more monomials is the product of the greatest common factor of the coefficients and the greatest common factors of the variables

Solved Examples on Monomial as a Common Factor

1. Factorize each of the following algebraic expressions.

(i) 15a + 10

Solution:
The given expression is 15a + 10
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 15 and 10 is 5.
HCF of literal coefficients:
The lowest power of a is 0.
Therefore, the HCF of literal coefficients is nothing.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 15a + 10 is 5.
Multiply and divide each term of the given expression 15a + 10 with 5a
5((15a/5) + (10/5)) = 5(3a + 2)

The final answer is 5(3a + 2)

(ii) 18mn² + 9m²n – 12mn

Solution:
The given expression is 18mn² + 9m²n – 12mn
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 18, 9, and 12 is 3.
HCF of literal coefficients:
The lowest power of m is 1.
The lowest power of n is 1.
Therefore, the HCF of literal coefficients is mn.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 18mn² + 9m²n – 12mn is 3mn.
Multiply and divide each term of the given expression 15a + 10 with 5a
3mn((18mn²/3mn) + (9m²n/3mn) – (12mn/3mn)) = 3mn (6n + 3m – 4)

The final answer is 3mn (6n + 3m – 4).

(iii) 20x²y – 8xy² + 32xy

Solution:
The given expression is 20x²y – 8xy² + 32xy
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 20, 8, and 32 is 4.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
Therefore, the HCF of literal coefficients is xy.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 20x²y – 8xy² + 32xy is 4xy.
Multiply and divide each term of the given expression 20x²y – 8xy² + 32xy with 4xy
4xy(20x²y/4xy – 8xy²/4xy+ 32xy/4xy) = 4xy (5x – 2y + 8)

The final answer is 4xy (5x – 2y + 8).

(iv) 13a³ + 39a²b

The given expression is 13a³ + 39a²b
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 13 and 39 is 13.
HCF of literal coefficients:
The lowest power of a is 2.
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is a².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 13a³ + 39a²b is 13a².
Multiply and divide each term of the given expression 13a³ + 39a²b with 13a².
13a²(13a³/13a² + 39a²b/13a²) = 13a² (a + 3b)

The final answer is 13a² (a + 3b).

Factors of Algebraic Expressions is the product of numbers, algebraic expressions, algebraic variables, etc. Here the numbers, algebraic expressions, algebraic variables are the factors of the algebraic expressions. Generally, factors of a number can be calculated with the product of its multiples. You can learn more about Factorization and related concepts by always seeking help from us. For example, by multiplying 2, 3, 7, we get 42. Therefore, 2, 3, 7 are the factors of 42. Factors of a given number are the product of two or more numbers.

Basics of Algebra Expressions

Expressions: Expressions are formed with constants and variables. 2x – 5 is an expression that formed with the variable x and numbers 2 and 5.

Terms, factors, and Coefficients: Algebraic expression is the combination of terms, coefficients, and factors. For example, in 5x + 7, 5x and 7 are terms, x and 7 are factors, and 5 and 7 are numeric coefficients.

Monomials, binomials, and Polynomials: An expression having only one term is known as a monomial, an expression having two terms are called binomial. Also, an expression with various terms and with a non-zero coefficient is called as Polynomials.

Factors of the Monomials

Product of variable and numbers can form a monomial that consists of different factors.

1. Write all the possible factors of 5ab²?

Solution:
The possible factors of 5 are 1 and 5.
The possible factors of ab² are a, b, b², ab, ab²
All possible factors of 5ab² are a, b, b², ab, ab², 1, 5, 5a, 5b, 5b², 5ab, and 5ab².

2. Write down all the factors of 7m²n?

Solution:
The possible factors of 7 are 1 and 7.
The possible factors of m²n are m, n, mn, m², m²n.
All possible factors of 7m²n are m, n, mn, m², m²n, 1, 7, 7m, 7n, 7mn, 7m², 7m²n.

3. Write all the factors of 3x²y²?

Solution:
The possible factors of 3 are 1 and 3.
The possible factors of x²y² are x, y, xy, x², y², x²y, xy², x²y²
All possible factors of 3x²y² are x, y, xy, x², y², x²y, xy², x²y², 1, 3, 3x, 3y, 3xy, 3x², 3y², 3x²y, 3xy², 3x²y²

4. Write down all the factors of 3xyz?

Solution:
The possible factors of 3 are 1 and 3.
The possible factors of xyz are x, y, z, xy, xz, yz, xyz
All possible factors of 3xyz are x, y, z, xy, xz, yz, xyz, 1, 3, 3x, 3y, 3z, 3xy, 3xz, 3yz, 3xyz

Highest Common Factor (HCF) of Monomials

The H.C.F. of two or more monomials is the product of the H.C.F. of the numerical coefficients and the common variables with the least powers.

1. Find the H.C.F. of 12a³b², 14a²b³, 6ab⁴.

Solution:
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 12, 14, and 6 is 2.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is ab².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 12a³b², 14a²b³, 6ab⁴ is 2ab².

The final answer is 2ab².

2. Find the H.C.F. of 6xy, 18x²y?

Solution:
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 6 and 18 is 6.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
Therefore, the HCF of literal coefficients is xy.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 6xy, 18x²y is 6xy.

The final answer is 6xy.

3. Find the H.C.F. of 9abc² and 36ac?

Solution:
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 9 and 36 is 9.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
The lowest power of c is 1.
Therefore, the HCF of literal coefficients is ac.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 9abc² and 36ac is 9ac.

The final answer is 9ac.

Factorization is nothing but the breaking down an entity (a number, a matrix, or a polynomial) into the product of another entity or factors. When you multiply the product of the factors, you will get the original number or matrix. Difficult algebraic or quadratic equations can reduce into small forms using the Factorization method. The factors of equations can be a variable, an integer, or an algebraic expression itself.

In simple words, Factorization can be explained as the reverse process of multiplication.

Examples:
(i) Product: 4x (3x – 6y) = 12x² – 24xy; Factorization: 12x² – 24xy = 4x (3x – 6y)
(ii) Product: (x + 4)(x – 3) = x² + x – 12; Factorization: x² + x – 12 = (x + 4)(x – 3
(iii) Product: (4a + 6b)(4a – 6b) = 16a² – 36b²; Factorization: 16a² – 36b² = (4a + 6b)(4a – 6b)

Maths Factorization

Find out different Factorization concepts and concern links below. You can learn each individual concept with a clear explanation with the help of the below links. We have explained every topic details separately with the solved examples. Therefore, students can easily get a grip on Factorization concepts by referring to our detailed concepts.

• Factors of Algebraic Expressions
• Monomial is a Common Factor
• Factorization when Monomial is Common
• Binomial is a Common Factor
• Factorization when Binomial is Common
• Factorization by Grouping
• Factorize by Grouping The Terms
• Factoring Terms by Grouping
• Factorization by Regrouping
• Factorize by Regrouping The Terms
• Factoring Terms by Regrouping
• Factorization by Using Identities
• Factorization of Perfect Square
• Factorization of Perfect Square Trinomials
• Difference of Two Squares
• Factoring Differences of Squares
• Factorize the Difference of Two Squares
• Evaluate the Difference of Two Squares
• Factorize the Trinomial x^2 + px + q
• Factorize the Trinomial ax^2 + bx + c
• Factorization of Quadratic Trinomials

Simple Factorization

Simple Factorization can be easily understood by the below examples.

(i) HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
(ii) HCF of literal coefficients = product of each common literal raised to the lowest power.

Factor

The factor is something that is to be multiplied.
Sum = term + term
Product = factor × factor
For example, x = m(n + 1); m and n + 1 are the factors.

Examples:

1. Factorize 4x²y² – 2xy

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 4 and 2 is 2.
HCF of literal coefficients:
The lowest power of x is 1
The lowest power of y is 1
Therefore, the HCF of literal coefficients is xy.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4x²y² – 2xy is 2xy.
Multiply and divide each term of the given expression 4x²y² – 2xy with 2xy
2xy((4x²y²/2xy) – (2xy/2xy)) = 2xy (2xy – 1)

The final answer is 2xy (2xy – 1)

2. Find the HCF of 24a³b²c³ and 27a⁴bc⁴.

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 24 and 27 is 3.
HCF of literal coefficients:
The lowest power of a is 3.
The lowest power of b is 1.
And, the lowest power of c is 3.
Therefore, the HCF of literal coefficients is a³bc³.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 24a³b²c³ and 27a⁴bc⁴ is 3a³bc³.

The final answer is 3a³bc³

3. Find the HCF of 2a²bc, 4a³b and 14bc.

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 2, 4, and 14 is 2.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 1.
And, the lowest power of c is 0.
Therefore, the HCF of literal coefficients is b.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 2a²bc, 4a³b and 14bc is 2b.

The final answer is 2b.

Factorization Solved Examples

1. Factorize 4y³ – 24y⁵

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 4 and 24 is 4.
HCF of literal coefficients:
The lowest power of y is 3
Therefore, the HCF of literal coefficients is y³.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4y³ – 24y⁵ is 4y³.
Multiply and divide each term of the given expression 4y³ – 24y⁵ with 4y³
4y³((4y³/4y³) – (24y⁵ /4y³) = 4y³(1 – 6y²)

The final answer is 4y³(1 – 6y²)

2. Factorize 21m²n⁵ – 7mn² + 28m⁵n

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 21, 7, and 28 is 7.
HCF of literal coefficients:
The lowest power of m is 1
The lowest power of n is 1
Therefore, the HCF of literal coefficients is mn.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 21m²n⁵ – 7mn² + 28m⁵n is 7mn.
Multiply and divide each term of the given expression 21m²n⁵ – 7mn² + 28m⁵n with 7mn
7mn((21m²n⁵ /7mn) – (7mn² /7mn) + (28m⁵n/7mn)) = 7mn(3mn⁴ – n + 4m⁴)

The final answer is 7mn(3mn⁴ – n + 4m⁴)

3. Factorize 3m(d + 5e) – 3n(d + 5e)

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 3 and 3 is 3.
HCF of literal coefficients:
The lowest power of m is 0
The lowest power of n is 0
Therefore, the HCF of literal coefficients is d+ 5e.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3m(d + 5e) – 3n(d + 5e) is 3(d + 5e).
Multiply and divide each term of the given expression 3m(d + 5e) – 3n(d + 5e) with 3(d + 5e)
3(d + 5e)((3m(d + 5e)/3(d + 5e)) – (3n(d + 5e)/3(d + 5e))) = 3(d + 5e)(m – n)

The final answer is 3(d + 5e)(m – n)

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Quadratic Equation is a second-degree polynomial equation with one variable x. The graph of quadratic equations makes nice curves. For every equation, we have two values of the variables called the roots. Get the formula and simple step by step process to solve the roots of any quadratic equation in the following sections.

Quadratic Equation Definition

The quadratic equation is an equation where the highest exponent of the variable is square. The standard form of quadratic equations is ax² + bx + c = 0. Where x is the variable, a, b, c are the constants and a should not be equal to zero. The power of x should not be negative and must be 2.

Quadratic Equation Formula

The formulas to find the solution or roots of the quadratic equation are given below:

(α, β) = [-b ± √(b² – 4ac)] / 2a

The values of the variable satisfying the given quadratic equation are called its roots. The roots of the quadratic equation details are mentioned here.

• Both roots of the equation are zero when b and c are zero.
• One of the roots of the quadratic equation is zero and the other is -b/a if c = 0
• The two roots are reciprocal to each other when the coefficients a and c are equal.

Nature of Roots of Quadratic Equation

In the quadratic equation formula, the term (b² – 4ac) is called the discriminant of the equation and it gives information about the nature of the roots. Below listed are the conditions that define the nature of roots.

• If the discriminant value is zero, then the equation will have equal roots i.e α = β = -b/2a.
• If the discriminant value is greater than zero, then the equation will have real roots.
• If the discriminant value is less than zero, then the equation will have imaginary roots.
• If the discriminant is greater than zero and it is a perfect square, then the equation will have rational roots.
• If the discriminant is greater than zero and it is not a perfect square, then the equation will have irrational roots.
• If the discriminant is greater than zero, perfect square, and a = 1, b, c are integers, then the equation will have integral roots.

How to Solve Quadratic Equations?

By solving the quadratic equations, you will get two roots that satisfy the equation. The easy steps to solve the quadratic equation roots are given here. The two ways to find the quadratic equations roots are the algebraic method and the graphical method. Here, we will learn about those methods.

Algebraic Method:

• Convert the given equation in the form of ax² + bx + c = 0, get a, b, c values.
• Substitute the values in the quadratic formula.
• Perform all the required calculations and find the roots.
• Another simple way is by factorizing the quadratic equation.
• The obtained factors are the roots.

Graphical Method:

• Plot the graph for the random values of x.
• The points where the curve meets the x-axis are the roots.

Examples of Quadratic Equations

(i) 5x² + 3x + 2 = 0 is a quadratic equation.

(ii) x – (1/x) = 6 is a quadratic equation

On solving this, we get x * x – 1 = 6 * x

x² – 1 = 6x

x² – 6x – 1 = 0

(iii) x² + √2x + 7 = 0 is not a quadratic equation.

(iv) x² + 5 = 0 is a quadratic equation.

(v) x² = 0,√ x² + 2x + 1 are quadratic equations.

Solved Examples on Finding Roots of the Equation

Example 1.

Solve the roots of equation 3x² – 22x – 16 by factorization?

Solution:

Given the quadratic equation is 3x² – 22x – 16

= 3x² – 24x + 2x – 16

= 3x(x – 8) + 2(x – 8)

= (x – 8) (3x + 2)

x – 8 = 0 and 3x + 2 = 0

x = 8 and 3x = -2, x = -2/3

The roots of the equation are α = 8, β = -2/3.

Therefore, solution set (8, -2/3)

Example 2.

Solve 9x² + 25x + 10 = 0 by using the quadratic formula.

Solution:

Given that,

9x² + 25x + 10 = 0

Roots of equation formula is (α, β) = [-b ± √(b² – 4ac)] / 2a

In the equation, a = 9, b = 25, and c = 10

Substitute these values in the equation.

α = [-b + √(b² – 4ac)] / 2a

= [-9 + √(9² – 4 (9) (10)] / 2 (9)

= [-9 + √(81 – 360)] / 18

= [-9 + √(-279)] / 18

= [-9 + √(279i²)] / 18

= [-9 + 16.70i] / 18

β = [-b – √(b² – 4ac)] / 2a

= [-9 – √(9² – 4 (9) (10)] / 2 (9)

= [-9 – √(81 – 360)] / 18

= [-9 – √(-279)] / 18

= [-9 – √(279i²)] / 18

= [-9 – 16.70i] / 18

Roots are α = [-9 + 16.70i] / 18, β = [-9 – 16.70i] / 18

Therefore, solution set ([-9 + 16.70i] / 18, [-9 – 16.70i] / 18)

Example 3.

Find the roots of the equation 1/(x + 4) – 1/(x – 7) = 11/30.

Solution:

Given that,

1/(x + 4) – 1/(x – 7) = 11/30

[(x – 7) – (x + 4)] / (x + 4) (x – 7) = 11/30

[x – 7 -x – 4] / (x² + 4x – 7x – 28) = 11/30

(-11) / (x² – 3x – 28) = 11/30

-1/ (x² – 3x – 28) = 1/30

-30 = x² – 3x – 28

x² – 3x – 28 + 30 = 0

x² – 3x + 2 = 0

x² – 2x – x + 2 = 0

x(x – 2) -1(x – 2) = 0

(x – 1) (x – 2) = 0

(x – 1) = 0 and (x – 2) = 0

x = 1, x = 2

The roots of quadratic equation is α = 1, β = 2.

Therefore, the solution set is (1, 2)

Example 4.

Find the values of a for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots.

Solution:

Given that,

(x – a) (x – 10) + 1 = 0

x² – ax – 10x + 10a + 1 = 0

x² – (a + 10)x + 10a + 1 = 0

Discriminant = b² – 4ac = (a + 10)² – 4 . 1. (10a + 1)

= a² + 20a + 100 – 40a – 4

= a² – 20a + 100 – 4

= (a – 10)² – 4

The quadratic equation will have integral roots, if the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers.

(a – 10)² – Discriminant = 4

Since discriminant is a perfect square. Hence, the difference of two perfect square in R.H.S will be 4 only when D = 0 and (a – 10)² = 4.

(a – 10) = ± 2

a = 2 + 10 or a = -2 + 10

= 12 and 8

Therefore, values of a are 8, 12.

Frequently Asked Questions on Quadratic Equations

1. What are the uses of the quadratic equations?

The quadratic equations can be used in calculating the area of rooms, speed of an object, determining product profit in the business. It can also be used in athletics and sports.

2. Write the differences between quadratic expression and quadratic equation?

The quadratic equation is a polynomial of degree 2 or any equation where quadratic expression is in the form of ax² + bx + c for equation or expression a ≠ 0. The quadratic formula is used to solve the roots of the equation.

3. What is the simplest form of a quadratic equation?

The standard form of a quadratic equation is ax² + bx + c – 0, where a ≠ 0. The coefficients a, b, c are real numbers.

4. Describe the nature of the roots?

The discriminant b² – 4ac > 0, roots are real, unequal and distinct. In case the discriminant = 0, roots are real, equal, and coincident. If the discriminant < 0, then roots are imaginary and unequal.

Practice Test on Linear Inequation has different types of questions. Students can test their skills and knowledge on linear inequations problems by solving all the provided questions on this page. The questions are mainly related to inequalities and finding the solution to the given inequation and draw a graph for the obtained solution set. You can easily draw a graph on a numbered line.

1. Write the equality obtained?

(i) On subtracting 1 from each side 3 > 7

(ii) On adding 3 to each side 12 < 5

(iii) On multiplying (-2) to each side 11 < 4

(iv) On multiplying 4 to each side 15 > 2

Solution:

(i) 3 – 1 > 7 – 1

2 > 6

(ii) 12 + 3 < 5 + 3

15 < 8

(iii) 11 x (-2)  4 x (-2)

-22 < -8

22 > 8

(iv) 15 x 4 > 2 x 4

60 > 8

2. Write the word statement for the following?

(i) x ≥ 15

(ii) x < 2

(iii) x ≤ -5

(iv) x > 16

(v) x ≠ 6

Solution:

(i) The variable x is greater than equal to 15. The possible values of x are 15 and more than 15.

(ii) The variable x is less than 2. The possible values of x are less than 2.

(iii) The variable x is less than and equal to -5. The possible values of x are less than -5.

(iv) The variable x is greater than 16. The possible values of x are more than 16.

(v) The variable x is not equal to 6. The possible values of x are all real numbers other than 6.

3. Find the solution set for each of the following inequations. x ∈ N

(i) x + 5 < 12

(ii) x – 6 > 5

(iii) 5x + 10 ≥ 17

(iv) 2x + 3 ≤ 6

Solution:

(i) x + 5 < 12

Subtract 5 from both sides.

x + 5 – 5 < 12 – 5

x < 7

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {1, 2, 3, 4, 5, 6}

(ii) x – 6 > 5

Add 6 to both sides.

x – 6 + 6 > 5 + 6

x > 11

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {12, 13, 14, 15, . . . }

(iii) 5x + 10 ≥ 17

Subtract 10 from both sides.

5x + 10 – 10 ≥ 17 – 10

5x ≥ 7

Divide 5 by each side.

5x/5 ≥ 7/5

x ≥ 1.4

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {2, 3, 4, 5 . . .}

(iv) 2x + 3 ≤ 6

Subtract 3 from both sides of the inequation

2x + 3 – 3 ≤ 6 – 3

2x ≤ 3

Both sides of the inequation divide by 2.

2x/2 ≤ 3/2

x ≤ 1.5

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {1, 1.5}

4. Find the solution set for each of the following inequations and represent it on the number line.

(i) 3 < x < 10, x ∈ N

(ii) 3x + 2 ≥ 6, x ∈ N

(iii) 3x/2 < 5, x ∈ N

(iv) -4 < 2x/3 + 1 < – 2, x ∈ N

Solution:

(i) 3 < x < 10, x ∈ N

The two cases are 3 < x and x < 10

It can also represent as x > 3 and x < 10

Replacement set = {1, 2, 3, 4, 5 . .}

The solution set for x > 3 is 4, 5, 6, 7 . . . i.e P = {4, 5, 6, 7 . . .}

And the solution set for x < 10 is 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e Q = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Therefore, solution set of the given inequation = P ∩ Q = {4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(ii) 3x + 2 ≥ 6, x ∈ N

Subtract 2 from both sides

3x + 2 – 2 ≥ 6 – 2

3x ≥ 4

Divide each side by 3

3x/3 ≥ 4/3

x ≥ 1.33

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {2, 3, 4, 5, . . }

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iii) 3x/2 < 5, x ∈ N

Multiply both sides by 2.

3x/2 x 2 < 5 x 2

3x < 10

divide both sides by 3

3x/3 < 10/3

x < 3.33

Replacement set = {1, 2, 3, 4, 5 . .}

Solution Set S = {1, 2}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iv) -4 < 2x/3 + 1 < – 2, x ∈ N

The two cases are -4 < 2x/3 + 1 and 2x/3 + 1 < – 2

Case I: -4 < 2x/3 + 1

Subtract 1 from both sides

-4 – 1 < 2x/3 + 1 – 1

-5 < 2x/3

Multiply each side by 3

-5 x 3 < 2x/3 x 3

-15 < 2x

Divide each side by 2

-15/2 < 2x/2

-7.5 < x

x > 7.5

Replacement Set = {1, 2, 3, 4, 5 . .}

Solution Set P = {8, 9, 10, 11 . . . }

Case II: 2x/3 + 1 < – 2

Subtract 1 from both sides

2x/3 + 1 – 1 < – 2 – 1

2x/3 < -3

Multiply 3 to both sides

2x/3 x 3 < -3 x 3

2x < -9

Divide both sides by 2

2x/2 < -9/2

x < -4.5

4.5 > x

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set Q = {1, 2, 3}

Therefore, required solution set S = P ∩ Q

S = Null

5. Find the solution set for each of the following and represent the solution set graphically?

(i) x – 6 < 4, x ∈ W

(ii) 6x + 2 ≤ 20, x ∈ W

(iii) 7x + 3 < 5x + 9, x ∈ W

(iv) 3x – 7 > 5x – 1, x ∈ I

Solution:

(i) x – 6 < 4, x ∈ W

Add 6 to both sides

x – 6 + 6 < 4 + 6

x < 10

Replacement set = {0, 1, 2, 3, 4, 5, 6, …}

Therefore, solution set S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(ii) 6x + 2 ≤ 20, x ∈ W

Subtract 2 from both sides

6x + 2 – 2 ≤ 20 – 2

6x ≤ 18

Divide each side by 6

6x/6 ≤ 18/6

x ≤ 3

Replacement set = {0, 1, 2, 3, 4, 5, 6, …}

Therefore, solution set S = {0, 1, 2, 3}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iii) 7x + 3 < 5x + 9, x ∈ W

Move variables to one side and constants to other side of inequation

7x – 5x < 9 – 3

2x < 6

Divide each side by 2

2x/2 < 6/2

x < 3

Replacement set = {0, 1, 2, 3, 4, 5, 6, …}

Therefore, solution set S = {0, 1, 2}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iv) 3x – 7 > 5x – 1, x ∈ I

Move variables to one side and constants to another side of inequation

-7 + 1 > 5x – 3x

-6 > 2x

divide 2 by each side

-6/2 > 2x/2

-3 > x

Replacement set ={ . . . -4, -3, -2, -1, 0, 1, 2, 3, . . .}

Solution set = { -2, -1, 0, 1, 2, . . . }

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.