Question

Find the product of -4p, 7pq.

Answer 1

Hint: We know that while calculating the product of two algebraic expressions we have to multiply the constants separately and variables separately. This will give us a product of two algebraic expressions. Let us assume the first algebraic expression is -4p and the second algebraic expression is 7pq. Let us assume the constant part of -4p is equal to \[{{C}_{1}}\]. In the similar way, let us assume the constant part of 7pq is equal to \[{{C}_{2}}\]. Now we have to find the product of \[{{C}_{1}}\] and \[{{C}_{2}}\].Let us assume this product as \[{{C}_{3}}\]. Let us assume the variable of -4p is equal to \[{{V}_{1}}\]. In the similar way, the constant part of 7pq is equal to \[{{V}_{2}}\]. Now we have to find the product of \[{{V}_{1}}\] and \[{{V}_{2}}\]. Let us assume this product as \[{{V}_{3}}\].Let us assume the product of given algebraic expressions is equal to C. Now we have to find the product of \[{{C}_{3}}\] and \[{{V}_{3}}\]. This gives us the algebraic expression C.

Complete step-by-step answer:
Before solving the question, we should know that while calculating the product of two algebraic expressions we have to multiply the constants separately and variables separately. This will give us a product of two algebraic expressions.
From the question, we were given that to find the product of -4p, 7pq. Now we have to divide the given algebraic expressions into two parts. Let us consider the first algebraic expression as A. From the given question, it was given that the first algebraic expression is -4p.
First algebraic expression:
\[\Rightarrow A=-4p…………...(1)\]
Now we have to divide the equation into two parts where first part represents a constant and the second part indicates variables.
In equation (1), -4 is the constant part and p is the variable part.
Let us assume the constant part is equal to \[{{C}_{1}}\].
\[\Rightarrow {{C}_{1}}=-4……………....(2)\]
Let us assume the variable part is equal to \[{{V}_{1}}\].
$\Rightarrow {{V}_{1}} = P $.................. (3)
Second algebraic expression:
\[\Rightarrow B=7pq ………………....(4)\]
Now we have to divide the equation into two parts where first part represents a constant and the second part indicates variables.
In equation (1), 7 is the constant part and pq is the variable part.
Let us assume the constant part is equal to \[{{C}_{2}}\].
\[\Rightarrow {{C}_{2}}=7…………………....(5)\]
Let us assume the variable part is equal to \[{{V}_{2}}\].
\[\Rightarrow {{V}_{2}}=pq ……………......(6)\]
Now to find the product of first algebraic expression and second algebraic expression, we have to find the product of constant parts of both algebraic expressions and product of variable parts of both algebraic expressions.
Let us assume the product of A and B is equal to C.
\[\Rightarrow C=AB ………….........(7)\]
Let us assume the constant part of C is equal to \[{{C}_{3}}\].
Now we have to find the product of \[{{C}_{1}}\] and \[{{C}_{2}}\].
\[\Rightarrow {{C}_{3}}={{C}_{1}}{{C}_{2}} ………………......(8)\]
Now we will substitute equation (2) and equation (5) in equation (8), we get
\[\begin{align}
  & \Rightarrow {{C}_{3}}=(-4)(7) \\
 & \Rightarrow {{C}_{3}}=-28 ………….........(9) \\
\end{align}\]
Let us assume the variable part of C is equal to \[{{V}_{3}}\].
Now we have to find the product of \[{{V}_{1}}\] and \[{{V}_{2}}\].
\[\Rightarrow {{V}_{3}}={{V}_{1}}{{V}_{2}}.....(10)\]
Now we will substitute equation (3) and equation (6) in equation (10), we get
\[\begin{align}
  & \Rightarrow {{V}_{3}}=(p)(pq) \\
 & \Rightarrow {{V}_{3}}={{p}^{2}}q …………..........(11) \\
\end{align}\]
We know that the product of constant part of an algebraic expression and variable part of an algebraic expression gives us the required algebraic expression.
In the similar way., the product of constant part of a C and variable part of a C gives us the algebraic expression for C.
\[\Rightarrow C={{C}_{3}}{{V}_{3}} ………........(12)\]
Now we have to substitute equation (9) and equation (11) in equation (12), we get
\[\begin{align}
  & \Rightarrow C=(-28)({{p}^{2}}q) \\
 & \Rightarrow C=-28{{p}^{2}}q ……………........(13) \\
\end{align}\]
From equation (13), it is clear that the product of -4p and 7pq is equal to \[-28{{p}^{2}}q\].

Note: Students may read the question in an incorrect manner. If a student read that it was given to find the product of 4p, 7pq. If a student considered this as the question, the solution as well as the final answer will get interrupted. So, students should read the question carefully.