Table of Contents

Last modified on January 10th, 2024

Multiplying monomials is a method of simplifying them by finding the product of their coefficients followed by the product of their variables.

For example, the product of two monomials, 5m and 7n, is:

5m ✕ 7n = (5 ✕ 7) ✕ (m ✕ n) = 35mn

Here, we observe that the product of monomials gives a monomial. Also, we usually write the variables in alphabetical order while writing the product.

Now, let us multiply another two monomials, 2p and 3p^{5}.

On multiplying the coefficients, we get

(2 ✕ 3) = 6

On multiplying the variables, we get

(p ✕ p^{5}), where both the variables have the same base and different exponents.

By applying the exponent rules (a^{m} ✕ a^{n} = a^{m+n}), we get

p^{1+5} = p^{6}

Thus, the product is 6p^{6}.

**Find the product of 3a ^{2}b^{3} and (4ab)^{3}**

Solution:

3a^{2}b^{3} ✕ (4ab)^{3}

Applying the exponent rules (a^{m} ✕ b^{m} = (ab)^{m}), we get

3a^{2}b^{3} ✕ 4^{3}a^{3}b^{3} = 3a^{2}b^{3} ✕ 64a^{3}b^{3}

Now, on multiplying the coefficients, we get

(3 ✕ 64) = 192

On multiplying the variables, we get

a^{2}b^{3} ✕ a^{3}b^{3} = (a^{2} ✕ a^{3}) ✕ (b^{3} ✕ b^{3})

As we know, a^{m} ✕ a^{n} = a^{m+n}, the product of the variables is

a^{2+3}b^{3+3}

Thus, the product of the given monomials is obtained as 192a^{5}b^{6}

To find the product of more than two monomials, we calculate the product of the first two monomials and then multiply it with the next monomial, and the process continues.

Let us multiply 3r^{4}s^{3}t, 2rs, and 5s^{2}t^{3}.

Now, on multiplying the coefficients, we get

(3 ✕ 2 ✕ 5) = 30

On multiplying the variables, we get

(r^{4}s^{3}t ✕ rs ✕ s^{2}t^{3}) = (r^{4} ✕ r) ✕ (s^{3} ✕ s ✕ s^{2}) ✕ (t ✕ t^{3}) = r^{1+4}s^{3+1+2}t^{1+3}

Thus, the product is r^{5}s^{6}t^{4}.

**Multiply: (3x ^{3})^{3}, 2x^{2}t, and 5x^{3}t^{6}.**

Solution:

(3x^{3})^{3} ✕ 2x^{2}t ✕ 5x^{3}t^{6}

Applying the exponent rules (a^{m} ✕ b^{m} = (ab)^{m} and (a^{m})^{n} = a^{mn}), we get

3^{3}(x^{3})^{3} ✕ 2x^{2}t ✕ 5x^{3}t^{6} = 27x^{9} ✕ 2x^{2}t ✕ 5x^{3}t^{6}

Simplifying further, we get

(27 ✕ 2 ✕ 5) ✕ (x^{9} ✕ x^{2} ✕ x^{3}) ✕ (t ✕ t^{6})

As we know, a^{m} ✕ a^{n} = a^{m+n}, the product of the given monomials is

270x^{9+2+3}t^{1+6} = 270x^{14}t^{7}.