Question

Find the product of $3{x^3}{y^2}$ and $\left( {2x - 3y} \right)$. Also, verify the result for $x = - 1,y = 2$.

Answer 1

Hint: Apply the method of the product of polynomial as when we find the product of any two polynomials, we just multiply each term of the first polynomial by each term of the second polynomial and then simplify to get the product. After that, substitute the value of $x,y$ in each polynomial and then multiply it and compare with the value of the product to verify it.

Complete step-by-step solution:
Given: - The polynomials are $3{x^3}{y^2}$ and $\left( {2x - 3y} \right)$.
Before proceeding with the solution, first, we will understand the concept of the product of one monomial and one polynomial. We will consider one monomial, $axy$ and one polynomial, $\left( {bx + cy} \right)$. To find the product of the monomial and polynomial, we have to multiply each term of the second polynomial by the monomial. So, the product of the monomial and polynomial is given as,
$axy \times \left( {bx + cy} \right) = ab{x^2}y + acx{y^2}$
Now, coming to the question, we are asked for the product of $3{x^3}{y^2}$ and $\left( {2x - 3y} \right)$. On multiplying the term, we get,
$ \Rightarrow 3{x^3}{y^2} \times \left( {2x - 3y} \right) = 3{x^3}{y^2} \times 2x - 3{x^3}{y^2} \times 3y$
Multiply the terms,
$ \Rightarrow 3{x^3}{y^2} \times \left( {2x - 3y} \right) = 6{x^4}{y^2} - 9{x^3}{y^3}$
So, the product is $6{x^4}{y^2} - 9{x^3}{y^3}$.
Now substitute the value $x = - 1,y = 2$ in each expression and get the value.
For $3{x^3}{y^2}$,
$ \Rightarrow 3{\left( { - 1} \right)^3}{\left( 2 \right)^2}$
Simplify the term,
$ \Rightarrow 3 \times - 1 \times 4$
Multiply the term,
$ \Rightarrow - 12$.................…… (1)
For $\left( {2x - 3y} \right)$,
$ \Rightarrow 2 \times - 1 - 3 \times 2$
Multiply the term,
$ \Rightarrow - 2 - 6$
Simplify the term,
$ \Rightarrow - 8$...............…… (2)
Now multiply equation (1) and (2),
$ \Rightarrow - 12 \times - 8 = 96$................….. (3)
For $6{x^4}{y^2} - 9{x^3}{y^3}$,
$ \Rightarrow 6{\left( { - 1} \right)^4}{\left( 2 \right)^2} - 9{\left( { - 1} \right)^3}{\left( 2 \right)^3}$
Simplify the term,
$ \Rightarrow 6 \times 1 \times 4 - 9 \times - 1 \times 8$
Multiply the term,
$ \Rightarrow 24 + 72$
Add the terms,
$ \Rightarrow 96$..................…… (4)
Now equate the equation (3) and equation (4),
$ \Rightarrow 96 = 96$

Hence, the product of $3{x^3}{y^2}$ and $\left( {2x - 3y} \right)$ is $6{x^4}{y^2} - 9{x^3}{y^3}$ and it is verified.

Note: While substituting the values of $x$ and $y$ in the expression, make sure that sign mistakes do not occur. Sign changes from + to - or from - to + when the number is shifted from one side of the equation to another.