
Complete the table of products
\[\begin{array}{*{20}{c}}
{\dfrac{{{\text{First monomial}}}}{{{\text{Second monomial}}}}}&{2x}&{ - 5y}&{3{x^2}}&{ - 4xy}&{7{x^2}y}&{ - 9{x^2}{y^2}} \\
{2x}&{4{x^2}}& - & - & - & - & - \\
{ - 5y}& - & - &{ - 15{x^2}y}& - & - & - \\
{3{x^2}}& - & - & - & - & - & - \\
{ - 4xy}& - & - & - & - & - & - \\
{7{x^2}y}& - & - & - & - & - & - \\
{ - 9{x^2}{y^2}}& - & - & - & - & - & -
\end{array}\]
Answer
587.4k+ views
Hint: In order to complete the table we are supposed to fill in the blank by multiplying it with the corresponding terms of the first and second monomials. The product of a term with itself gives a square of that number, and similarly cube, fourth power and so on.
Complete step-by-step answer:
Let us see some of the examples one by one and further solve the table.
For filling the second row and third column we have to multiply $2x$ and $ - 5y$ .
So the product is given as:
$
= 2x \times - 5y \\
= - 10xy \\
$
Similarly if we have the second row and the fourth column we have to multiply $2x$ and $3{x^2}$ .
The multiplication of each of the terms is given as,
So the product is given as:
$
= 2x \times 3{x^2} \\
= 6{x^3} \\
$
Similarly we fill out each and every blank of the table. Hence it looks like
\[\begin{array}{*{20}{c}}
{\dfrac{{{\text{First monomial}}}}{{{\text{Second monomial}}}}}&{2x}&{ - 5y}&{3{x^2}}&{ - 4xy}&{7{x^2}y}&{ - 9{x^2}{y^2}} \\
{2x}&{4{x^2}}&{ - 10xy}&{6{x^3}}&{ - 8{x^2}y}&{14{x^3}y}&{ - 18{x^3}{y^2}} \\
{ - 5y}&{ - 10xy}&{25{y^2}}&{ - 15{x^2}y}&{20x{y^2}}&{ - 35{x^2}{y^2}}&{ - 45{x^2}{y^3}} \\
{3{x^2}}&{6{x^3}}&{ - 15{x^2}y}&{9{x^4}}&{ - 12{x^3}y}&{21{x^4}y}&{ - 27{x^4}{y^2}} \\
{ - 4xy}&{ - 8{x^2}y}&{20x{y^2}}&{ - 12{x^3}y}&{16{x^2}{y^2}}&{ - 28{x^3}{y^2}}&{36{x^3}{y^3}} \\
{7{x^2}y}&{14{x^3}y}&{ - 35{x^2}{y^2}}&{21{x^4}y}&{ - 28{x^3}{y^2}}&{49{x^4}{y^2}}&{ - 63{x^4}{y^3}} \\
{ - 9{x^2}{y^2}}&{ - 18{x^3}{y^2}}&{ - 45{x^2}{y^3}}&{ - 27{x^4}{y^2}}&{36{x^3}{y^3}}&{ - 63{x^4}{y^3}}&{81{x^4}{y^4}}
\end{array}\]
Hence the answer is given in the table.
Note: In order to solve problems of this type the key is to identify that the blank is to be filled with the product of the corresponding terms by observing the given. In mathematics, a monomial is a polynomial with one term. When you multiply monomials, first multiply the coefficients and then multiply the variables by adding the exponents. Note that when you multiply monomials with the same base, you can add their exponents. This is called the Product of Powers Property.
Complete step-by-step answer:
Let us see some of the examples one by one and further solve the table.
For filling the second row and third column we have to multiply $2x$ and $ - 5y$ .
So the product is given as:
$
= 2x \times - 5y \\
= - 10xy \\
$
Similarly if we have the second row and the fourth column we have to multiply $2x$ and $3{x^2}$ .
The multiplication of each of the terms is given as,
So the product is given as:
$
= 2x \times 3{x^2} \\
= 6{x^3} \\
$
Similarly we fill out each and every blank of the table. Hence it looks like
\[\begin{array}{*{20}{c}}
{\dfrac{{{\text{First monomial}}}}{{{\text{Second monomial}}}}}&{2x}&{ - 5y}&{3{x^2}}&{ - 4xy}&{7{x^2}y}&{ - 9{x^2}{y^2}} \\
{2x}&{4{x^2}}&{ - 10xy}&{6{x^3}}&{ - 8{x^2}y}&{14{x^3}y}&{ - 18{x^3}{y^2}} \\
{ - 5y}&{ - 10xy}&{25{y^2}}&{ - 15{x^2}y}&{20x{y^2}}&{ - 35{x^2}{y^2}}&{ - 45{x^2}{y^3}} \\
{3{x^2}}&{6{x^3}}&{ - 15{x^2}y}&{9{x^4}}&{ - 12{x^3}y}&{21{x^4}y}&{ - 27{x^4}{y^2}} \\
{ - 4xy}&{ - 8{x^2}y}&{20x{y^2}}&{ - 12{x^3}y}&{16{x^2}{y^2}}&{ - 28{x^3}{y^2}}&{36{x^3}{y^3}} \\
{7{x^2}y}&{14{x^3}y}&{ - 35{x^2}{y^2}}&{21{x^4}y}&{ - 28{x^3}{y^2}}&{49{x^4}{y^2}}&{ - 63{x^4}{y^3}} \\
{ - 9{x^2}{y^2}}&{ - 18{x^3}{y^2}}&{ - 45{x^2}{y^3}}&{ - 27{x^4}{y^2}}&{36{x^3}{y^3}}&{ - 63{x^4}{y^3}}&{81{x^4}{y^4}}
\end{array}\]
Hence the answer is given in the table.
Note: In order to solve problems of this type the key is to identify that the blank is to be filled with the product of the corresponding terms by observing the given. In mathematics, a monomial is a polynomial with one term. When you multiply monomials, first multiply the coefficients and then multiply the variables by adding the exponents. Note that when you multiply monomials with the same base, you can add their exponents. This is called the Product of Powers Property.
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