# Worksheet on Factoring Algebraic Expression | Factoring Algebraic Expressions Worksheet

Take the help of Worksheet on Factoring Algebraic Expression while preparing for the exam. You can have better practice with the Factorization Worksheets given here. A variety of questions along with answers are included along with an explanation. Factoring Algebraic Expressions Worksheets are developed as per the latest syllabus. Therefore, students can follow the questions given in Factorization Worksheets and get good marks in the exam.

1. Write all the possible factors of

(a) 4a
(b) 10ab
(c) 12mn2
(d) 24m2n
(e) 5ab2c2

Solution:

(a) Given expression is 4a.
The possible factors of 4 are 1, 2, and 4.
The possible factors of a is a.

All possible factors of 4a are 1, a, 2, 2a, 4, 4a.

(b) Given expression is 10ab.
The possible factors of 10 are 1, 2, 5, and 10.
The possible factors of ab is a, b, and ab.

All possible factors of 10ab are 1, 2, 5, and 10, a, b, ab, 2a, 2b, 2ab, 5a, 5b, 5ab, 10a, 10b, 10a.

(c) Given expression is 12mn2.
The possible factors of 12 are 1, 2, 3, 4, 6, and 12.
The possible factors of mn2 are m, n, n2, mn2, mn.

All possible factors of 12mn2 are 1, 2, 3, 4, 6, 12, m, n, n2, mn2, mn, 2m, 2n, 2n2, 2mn, 2mn2, 3m, 3n, 3n2, 3mn, 3mn2, 4m, 4n, 4n2, 4mn, 4mn2, 6m, 6n, 6n2, 6mn, 6mn2, 12m, 12n, 12n2, 12mn and 12mn2

(d) Given expression is 24m2n.
The possible factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
The possible factors of m2n are m, m2, n, mn, m2n.

All possible factors of 24m2n are 1, 2, 3, 4, 6, 8, 12, 24, m, m2, n, mn, m2n, 2m, 2m2, 2n, 2mn, 2m2n, 3m, 3m2, 3n, 3mn, 3m2n, 4m, 4m2, 4n, 4mn, 4m2n, 6m, 6m2, 6n, 6mn, 6m2n, 8m, 8m2, 8n, 8mn, 8m2n, 12m, 12m2, 12n, 12mn, 12m2n, 24m, 24m2, 24n, 24mn, 24m2n.

(e) Given expression is 5ab2c2.
The possible factors of 5 are 1, and 5.
The possible factors of ab2c2 are a, b, b2, c, c2, ab, ab2, bc, bc2, ac, ac2, b2c2.

All possible factors of 5ab2c2 are 1, 5, a, b, b2, c, c2, ab, ab2, bc, bc2, ac, ac2, b2c2, 5a, 5b, 5b2, 5c, 5c2, 5ab, 5ab2, 5bc, 5bc2, 5ac, 5ac2, 5b2c2.

2. Find the H.C.F. of the monomials

(a) 4a2 and 10ab
(b) 21m2n and 49mn2
(c) a3b2 and – 5b2
(d) 4a3, 6b2 and 8c
(e) 2a2b3, 10a3b2 and 14ab
(f) 5m3, -15m2 and 45m

Solution:

(a) Given that 4a2 and 10ab
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4 and 10 is 2.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4a2 and 10ab is 2a.

(b) Given that 21m2n and 49mn2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 21 and 49 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 a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 21m2n and 49mn2 is 7mn.

(c) Given that a3b2 and – 5b2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1 and -5 is 1.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is b2.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of a3b2 and – 5b2 is b2.

(d) Given that 4a3, 6b2 and 8c
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4, 6, and 8 is 2.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is 0.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4a3, 6b2 and 8c is 2.

(e) Given that 2a2b3, 10a3b2 and 14ab
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 2, 10, and 14 is 2.
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 2a2b3, 10a3b2 and 14ab is 2ab.

(f) Given that 5m3, -15m2, and 45m
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 5, -15, and 45 is 5.
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 5m3, -15m2, and 45m is 5m.

3. Find the common factors of the given terms.

(a) 3a2, 15ab
(b) 33a2b, – 77ab2
(c) 9a2b2c, 54abc2
(d) a3b2, – 5b2
(e) 4a3, 10b3, 6c3
(f) 2abc3, 3a2b2c, 5abc

Solution:

(a) Given that 3a2, 15ab
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3 and 15 is 3.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3a2 and 15ab is 3a.

(b) Given that 33a2b, – 77ab2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 33 and -77 is 11.
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 33a2b, and – 77ab2 is 11ab.

(c) Given that 9a2b2c, 54abc2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 9 and 54 is 9.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
The lowest power of c is 1.
Therefore, the HCF of literal coefficients is abc.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 9a2b2c and 54abc2 is 9abc.

(d) Given that a3b2, – 5b2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1 and -5 is 1.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is b2.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of a3b2 and – 5b2 is b2.

(e) Given that 4a3, 10b3, and 6c3
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4, 10, and 6 is 2.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is 1.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4a3, 10b3, and 6c3 is 2.

(f) Given that 2abc3, 3a2b2c, 5abc
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 2, 3, and 5 is 1.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
The lowest power of c is 1.
Therefore, the HCF of literal coefficients is abc.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 2abc3, 3a2b2c, and 5abc is abc.

4. Find the H.C.F. of the terms of the expression

(a) – 2a2 – 8a3 – 20a4
(b) 3mn2 + 9m2n2 – 6m2n
(c) 5a2 + 25a + 50
(d) a3b2 – 8b2

Solution:

(a) Given that – 2a2 – 8a3 – 20a4
Firstly, find the HCF of given terms.
HCF of their numerical coefficients -2, -8, and -20 is -2.
HCF of literal coefficients:
The lowest power of a is 2.
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 – 2a2 – 8a3 – 20a4 is -2a2.

(b) Given that 3mn2 + 9m2n2 – 6m2n
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, 9, and -6 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 3mn2 + 9m2n2 – 6m2n is 3mn.

(c) Given that 5a2 + 25a + 50
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 5, 25, and 50 is 5.
HCF of literal coefficients:
The lowest power of a is 0.
Therefore, the HCF of literal coefficients is 1.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 5a2 + 25a + 50 is 5.

(d) Given that a3b2 – 8b2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1, and -8 is 1.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is b2.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of a3b2 – 8b2 is b2.

5. Find the factors of the expression

(a) 3a4 – 9a5 – 15a3
(b) – 10mn3 + 30nm3 – 50m2n3
(c) 15x2 – 20y2 + 25z2
(d) a3bc + 5ab3+ 9a3b
(e) 3a2 – 9a2b – 27a3c
(f) 7a3 + 7ab2 + 7ac2

Solution:

(a) The given expression is 3a4 – 9a5 – 15a3
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, -9, and -15 is 3.
HCF of literal coefficients:
The lowest power of a is 3.
Therefore, the HCF of literal coefficients is a3.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3a4 – 9a5 – 15a3 is 3a3.
Multiply and divide each term of the given expression 3a4 – 9a5 – 15a3 with 3a3.
3a3 (3a4/3a3 – 9a5/3a3 – 15a3/3a3) = 3a3 (a – 3a2 – 5)

The final answer is 3a3 (a – 3a2 – 5).

(b) The given expression is – 10mn3 + 30nm3 – 50m2n3
Firstly, find the HCF of given terms.
HCF of their numerical coefficients -10, 30, and -50 is -10.
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 – 10mn3 + 30nm3 – 50m2n3 is -10mn.
Multiply and divide each term of the given expression – 10mn3 + 30nm3 – 50m2n3 with -10mn.
-10mn (– 10mn3/-10mn+ 30nm3/-10mn – 50m2n3/-10mn) = -10mn (n2 – 3m2 + 5mn2)

The final answer is -10mn (n2 – 3m2 + 5mn2).

(c) The given expression is 15x2 – 20y2 + 25z2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 15, -20, and 25 is 5.
HCF of literal coefficients:
The lowest power of x is 0.
The lowest power of y is 0.
The lowest power of z is 0.
Therefore, the HCF of literal coefficients is 1.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 15x2 – 20y2 + 25z2 is 5.
Multiply and divide each term of the given expression 15x2 – 20y2 + 25z2 with 5.
5 (15x2/5 – 20y2/5 + 25z2/5) = 5 (3x2 – 4y2 + 5z2)

The final answer is 5 (3x2 – 4y2 + 5z2).

(d) The given expression is a3bc + 5ab3+ 9a3b
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1, 5, and 9 is 1.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
The lowest power of c is 0.
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 a3bc + 5ab3+ 9a3b is ab.
Multiply and divide each term of the given expression a3bc + 5ab3+ 9a3b with ab.
ab (a3bc/ab + 5ab3/ab + 9a3b/ab) = ab (a2c + 5b2 + 9a2)

The final answer is ab (a2c + 5b2 + 9a2).

(e) The given expression is 3a2 – 9a2b – 27a3c
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, -9, and -27 is 3.
HCF of literal coefficients:
The lowest power of a is 2.
The lowest power of b is 0.
The lowest power of c 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 3a2 – 9a2b – 27a3c is 3a2.
Multiply and divide each term of the given expression 3a2 – 9a2b – 27a3c with 3a2.
3a2 (3a2/3a2 – 9a2b/3a2 – 27a3c/3a2) = 3a2 (1 – 3b – 9ac)

The final answer is 3a2 (1 – 3b – 9ac).

(e) The given expression is 3a2 – 9a2b – 27a3c
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, -9, and -27 is 3.
HCF of literal coefficients:
The lowest power of a is 2.
The lowest power of b is 0.
The lowest power of c 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 3a2 – 9a2b – 27a3c is 3a2.
Multiply and divide each term of the given expression 3a2 – 9a2b – 27a3c with 3a2.
3a2 (3a2/3a2 – 9a2b/3a2 – 27a3c/3a2) = 3a2 (1 – 3b – 9ac)

The final answer is 3a2 (1 – 3b – 9ac).

(f) The given expression is 7a3 + 7ab2 + 7ac2
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 7, 7, and 7 is 7.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is 7a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 7a3 + 7ab2 + 7ac2 is 7a.
Multiply and divide each term of the given expression 7a3 + 7ab2 + 7ac2 with 7a.
7a (7a3/7a + 7ab2/7a + 7ac2/7a) = 7a (a2 + b2 + c2)

The final answer is 7a (a2 + b2 + c2).