Practice using the Worksheet on Dividing Monomials and improve your algebra basics. You can solve different types of algebraic expressions much easily provided you have a grip on the Division of Monomials Concept. All the Problems in the Dividing Monomials Worksheet are designed with a simple approach so that you will become familiar with the concept step by step. The Free Printable Math Worksheet on Division of Monomials available in PDF Formats can be accessed for free of cost and you can practice regularly.

See More:

• Worksheet on Addition and Subtraction of Polynomials
• Worksheet on Multiplying Monomial and Binomial
• Worksheet on Multiplying Monomial and Polynomial

## Dividing Monomials Worksheet with Answers

I. Divide the following monomials and write the answer in simplest form:
(i) (48xy) ÷ (4x)
(ii) (12p6q4) ÷ (-6p2q2)
(iii) (26m2pn2) ÷ (-2mn)
(iv) (-96x4y3) ÷ (-8x2y)
(v) (-56b2x3y5) ÷ (4bx2y2)
(vi) (48ab) ÷ (-4)
(vii) (75m2n) ÷ (5n)

Solution:

(i) Given monomials are 48xy,4x
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
48xy/4x=6 × 8 × x × y/2 × 2 × x
=24y/2
=12y
Therefore, By dividing (48xy) with (4x) we get 12y.

(ii) Given (12p6q4) ÷ (-6p2q2)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
(12p6q4) / (-6p2q2)=2 × 6 × p × p ×p ×p × p ×p × q × q × q × q/-6 × p × p × q × q
=-2p4q2
Therefore, By dividing 12p6q4 with -6p2q2 we get -2p4q2.
(iii) Given (26m2pn2) ÷ (-2mn)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
26m2pn2/-2mn=2 × 13 × m × m × p × n × n/-2 ×m ×n
=-13mpn
Therefore, By dividing 26m2pn2 with -2mn we get -13mpn.

(iv) Given (-96x4y3) ÷ (-8x2y)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
-96x4y3/8x2y=-12 × 8 × x × x × x × x × y × y × y/8 × x × x ×y
=-12 x2y2
Therefore, By dividing -96x4y3 with-8x2y we get -12 x2y2.
(v) Given (-56 b2x3y5) ÷ (4bx2y2)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
-56 b2x3y5/4bx2y2= -4 × 14 × b × b × x ×  x × x × y × y × y × y × y/4 × b × x ×  x ×  y ×  y
=-14bxy3
Therefore, By dividing -54b2x3y5 with 4bx2y2 we get -14bxy3.
(vi) Given (48ab) ÷ (-4)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
48ab/-4=12 × 4 × a × b/-4
=-12ab
Therefore, By dividing 48ab with -4 we get -12ab.
(vii) Given (75m2n) ÷ (5n)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
75m2n/5n= 15 × 5 × m × m × n/5 × n
=15m2
Therefore, By dividing 75m2n with 5n we get 15m2 .

II. Divide a monomial by a monomial:
(i) 60a2 ÷ 5a
(ii) 63x2y2 ÷ 7xy
(iii) 48a4b6 ÷ 12a2b2
(iv) 108m4n3 ÷ (-18m2n2)
(v) (-46a4b5c2) ÷ (-23a2bc)

Solution:

(i) Given, 60a2 ÷ 5a
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
60a2/5a= 12 × 5 × a × a/5 × a
=12a
Hence, By dividing 60a2 with 5a we get 12a.
(ii) Given, 63x2y2 ÷ 7xy
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
63x2y2/7xy=7 × 9 × x × x × y× y/7 × x × y
=9xy
Therefore, By dividing 63x2y2 with 7xy we get 9xy.
(iii) Given, 48a4b6 ÷ 12a2b2
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
48a4b6/12a2b2= 12 × 4 × a × a × a × a × b × b × b × b × b × b/4 × 3 × a × a × b × b
=4a2b4
Therefore, By dividing 48a4b6 with 12a2b2 we get 4a2b4.
(iv) Given, 108m4n3 ÷ (-18m2n2)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
108m4n3/-18m2n2= 18 × 6 × m × m × m × m  × n × n × n/-6 × 3 × m × m × n × n
=-6m2n
Therefore, By dividing 108m4n with -18m2n2 we get -6m2n.
(v) Given, (-46a4b5c2) ÷ (-23a2bc)
Now we have to write each term in expanded form and then cancel the terms which are common to both numerator and denominator.
-46a4b5c2/-23a2bc= -23 × 2 × a × a × a × a × b × b × b × b × b × c × c/-23 × a × a × b × b × c
=2a2b4c
Hence, By dividing -46a4b5c2 with -23a2bc we get 2a2b4c.

III. Divide first monomial by the second monomial:
(i) 30xy, 6x
(ii) 75a4b, 5b
(iii) 48ab4c, 8b2c
(iv) 30a5, (-a2)
(v) 22a2b5, 2ab

Solution:

(i) Given 30xy, 6x
30xy/6x=5y
Hence, By dividing the first monomial 30xy with the second monomial 6x we get 5y.
(ii) Given 75a4, 5b
75a4b/5b=15a4
Therefore, By dividing the first monomial 75a4b with the second monomial 5b we get 15a4.
(iii) Given 48ab4c, 8b2c
48ab4c/8b2c=6ab2
Therefore, By dividing the first monomial 48ab4c with the second monomial 8b2c we get 6ab2.
(iv) Given 30a5, (-a2)
30a55/-a2=-30a3
Hence, By dividing the first monomial 30a5 with the second monomial -a2 we get -30a3.
(v) Given 22a2b5, 2ab
22a2b5/ 2ab=11ab4
Therefore, By dividing the first monomial 22a2b5 with the second monomial 2ab we get 11ab4.

IV. Simplify the division of the monomials
(i) (-18x8) ÷ (-9x2)
(ii) (40a4b8c4) ÷ (5a2b2c2)
(iii) (14x4y3z2) ÷ (-2x2yz)
(iv) (81a7b8c3) ÷ (9a4b33c3)
(v) (51x6y5z4) ÷ (-3x4y2z)

Solution:

(i) Given (-18×8) ÷ (-9x2)
-18x8/-9x2=2x6
Hence, By dividing -18x8 with -9x2 we get 2x6.
(ii) Given (40a4b8c4) ÷ (5a2b2c2)
40a4b8c4/5a2b2c2=8a2b6c2
Hence, By dividing 40a44b8c4 with 5a2b2c2 we get 8a2b6c2.
(iii) Given (14x44y3z2) ÷ (-2x2yz)
14x4y3z2/-2x2yz=-7x2y2z
Hence, By dividing 14x4y3z2 with -2x2yz we get -7x2y2z.
(iv) Given (81a7b8c3) ÷ (9a4b3c3)
81a7b8c3/9a4b3c3=9a3b5
Hence, By dividing 81a7b8c3 with 9a4b3c3 we get 9a3b5.
(v) Given (51x6y5z4) ÷ (-3x4y2z)
51x6y5z4/-3x4y2z=-17x22y3z3
Hence, By dividing 51x6y5z4 with -3x4y2z we get -17x22y3z3.