Question

Find the product of the following pairs of monomials.
$
  \left( i \right)4,7p \\
  \left( {ii} \right) - 4p,7p \\
  \left( {iii} \right) - 4p,7pq \\
  \left( {iv} \right)4{p^3}, - 3p \\
  \left( v \right)4p,0 \\
$

Answer 1

Hint: When terms with similar variables (exponent>0) irrespective of their exponents are multiplied, then the exponent of the result must be greater than the multiplicands. Anything multiplied with zero is zero.

Complete step-by-step solution:
$\left( i \right)4,7p$
Here, ‘4’ is a constant with degree 0 and ‘7p’ is a monomial with degree 1.
‘7p’ has a constant ‘7’ and a variable ‘p’.
‘4’ can be written in terms of ‘p’ as $4 \times {p^0}$
‘7p’ can be written as $7 \times {p^1}$
$
  4 \times 7p = 4 \times {p^0} \times 7 \times {p^1} \\
   = \left( {4 \times 7} \right) \times \left( {{p^0} \times {p^1}} \right) \\
   = 28 \times {p^{0 + 1}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = 28{p^1} \\
   = 28p \\
    \\ $
Therefore, the product of $4,7p$ is $28p$
$\left( {ii} \right) - 4p,7p$
Here both the terms ‘-4p’, ‘7p’ are monomials with degree 1 and the same variable ‘p’.
‘-4p’ has an integer ‘-4’ and a variable ‘p’.
‘7p’ has an integer ‘7’ and a variable ‘p’.
‘-4p’ can be written as $ - 4 \times {p^1}$
‘7p’ can be written as $7 \times {p^1}$
$
   - 4p \times 7p = - 4 \times {p^1} \times 7 \times {p^1} \\
   = - 4 \times 7 \times {p^1} \times {p^1} \\
   = - 28 \times {p^{1 + 1}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = - 28{p^2} \\
$
Therefore, the product of $ - 4p,7p$ is $ - 28{p^2}$
$\left( {iii} \right) - 4p,7pq$
Here both terms ‘-4p’, ‘7pq’ are monomials with degree 1 and 2 respectively.
‘-4p’ has an integer ‘-4’ and a variable ‘p’.
‘7pq’ has an integer ‘7’ and variables ‘p’, ‘q’.
‘-4p’ can be written as $ - 4 \times {p^1} \times {q^0}$
‘7pq’ can be written as $7 \times {p^1} \times {q^1}$
$
   - 4p \times 7pq = - 4 \times {p^1} \times {q^0} \times 7 \times {p^1} \times {q^1} \\
   = - 4 \times 7 \times {p^1} \times {p^1} \times {q^0} \times {q^1} \\
   = - 28 \times {p^{1 + 1}} \times {q^{0 + 1}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = - 28 \times {p^2} \times {q^1} \\
   = - 28{p^2}q \\
$
Therefore, the product of $ - 4p,7pq$ is $ - 28{p^2}q$
$\left( {iv} \right)4{p^3}, - 3p$
Here both terms are monomials with degree 3 and degree 1 respectively.
Both the monomials have an integer and a similar variable ‘p’.
‘-3p’ can be written as $ - 3 \times {p^1}$
$
  4{p^3} \times - 3p = 4 \times {p^3} \times \left( { - 3} \right) \times {p^1} \\
   = 4 \times \left( { - 3} \right) \times {p^3} \times {p^1} \\
   = - 12 \times {p^{3 + 1}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = - 12 \times {p^4} \\
   = - 12{p^4} \\
$
Therefore, the product of $4{p^3}, - 3p$ is $ - 12{p^4}$
$\left( v \right)4p,0$
Here both the terms are monomial.
‘4p’ has an integer ‘4’ and a variable ‘p’ with degree 1.
And the second term 0(zero) is an integer.
As we know that, any term or any variable or any number multiplied with zero results zero.
So, the product of $4p,0$ is $0$

Note: The degree of a monomial is defined as the sum of all the exponents of the variables, including the implicit exponents of 1 for the variables which appear without exponent. The degree of all the constants is zero.