Table of Contents

Last modified on January 10th, 2024

Factoring monomials is a way of breaking the monomial into smaller terms and expressing them in the product form.

For example, the monomial 6m^{3} is factored as:

- (2)(3m
^{3}) = 6m^{3} - (3)(2m
^{3}) = 6m^{3} - (2m)(3m
^{2}) = 6m^{3} - (2m
^{2})(3m) = 6m^{3} - (6)(m
^{3}) = 6m^{3} - (6m)(m
^{2}) = 6m^{3} - (6m
^{2})(m) = 6m^{3} - (6m
^{3})(1) = 6m^{3}

Also, we observe that the product of each of the factors gives back the original monomial 6m^{3}. Thus, the possible factors of the monomial 6m^{3} are 1, 2, 3, 6, m, m^{2}, m^{3}, 2m, 2m^{2}, 2m^{3}, 3m, 3m^{2}, 3m^{3}, 6m, 6m^{2}, 6m^{3}.

However, finding factors with this method takes time. Instead, we expand the monomials completely by factoring like prime factorization of integers.

While factoring a monomial completely, we expand each part of the monomial until it is no longer expandable.

Again, taking the monomial 6m^{3} and completely factoring it, we get

6m^{3} = 2 Ã— 3 Ã— m Ã— m Ã— m

Thus, 2 Ã— 3 Ã— m Ã— m Ã— m is the complete factorization of 6m^{3}.

**What are the factors of 35v?**

Solution:

35v = (1)(35v), (5)(7v), (7)(5v), (35)(v).

Thus, the factors of 35v are 1, 5, 7, 35, v, 5v, 7v, 35v.

**Factor the monomial 16x ^{2}y completely.**

Solution:

16x^{2}y = 2 Ã— 2 Ã— 2 Ã— 2 Ã— x Ã— x Ã— y

Thus, 2 Ã— 2 Ã— 2 Ã— 2 Ã— x Ã— x Ã— y is the complete factorization of 16x^{2}y.

The greatest common factor (GCF) of monomials, also called the common monomial factor, refers to the product of the greatest common factor of coefficients and the greatest common factor of the variables.

To find the GCF of two or more monomials, we factor them completely and then find the product of all the common factors.

Now, let us find the GCF of 21p^{4}q^{2} and -14q^{5}p^{3}.

Factoring them completely, we get

21p^{4}q^{2} = 3 Ã— 7 Ã— p Ã— p Ã— p Ã— p Ã— q Ã— q

-14q^{5}p^{3} = -1 Ã— 2 Ã— 7 Ã— q Ã— q Ã— q Ã— q Ã— q Ã— p Ã— p Ã— p

Thus, the GCF is 7 Ã— p Ã— p Ã— p Ã— q Ã— q.

**Find the GCF of 36m ^{2}n^{2}, 20mn, and 40mn^{2}.**

Solution:

Factoring the given monomials completely, we get

36m^{2}n^{2} = 2 Ã— 2 Ã— 3 Ã— 3 Ã— m Ã— m Ã— n Ã— n

20mn = 2 Ã— 2 Ã— 5 Ã— m Ã— n

40mn^{2} = 2 Ã— 2 Ã— 2 Ã— 5 Ã— m Ã— n Ã— n

Thus, the GCF of 36m^{2}n^{2}, 20mn, and 40mn^{2} = 2 Ã— 2 Ã— m Ã— n = 4mn.

Last modified on January 10th, 2024