Question

Complete the table of the product by multiplying the first monomial present along the row with the second monomial present along the column.

$\begin{array}{ccccc}Nos.& FirstMonomial\to secondMonomial\downarrow & a^2{b}^2& c{d}^2& -5p^2\\ \left(i\right)& -ab& -a^3{b}^3& -abc{d}^2& \\ \left(ii\right)& -cd& -a^2{b}^{2}cd& -c^2{d}^3& \\ \left(iii\right)& abcd& a^3{b}^{3}cd& ab{c}^2{d}^3& \end{array}$

• A. 5abp², 5cdp², -5abcdp²
• B. -5abp³, 5c²dp², -5abcdp²
• C. 5abp², 5cdp², 5a²b²cdp²
• D. 5a²bp², 5c²dp², -5a³b³cdp²

Answer 1

The correct option is A

5abp², 5cdp², -5abcdp²

(i). (-ab) $\times$ (-5p${}^2$) = 5abp${}^2$

(ii). (-cd) $\times$ (-5p${}^2$) = 5cdp${}^2$

(iii). (abcd) $\times$ (-5p${}^2$) = -5 abcdp${}^2$

$\begin{array}{cccccc}Nos.& FirstMonomial\to & SecondMonomial\downarrow & a^2{b}^2& c{d}^2& -5p^2\\ \left(i\right)& -ab& -a^3{b}^3& -abc{d}^2& 5ab{p}^2& \\ \left(ii\right)& -cd& -a^2{b}^{2}cd& -c^2{d}^3& 5cd{p}^2& \\ \left(iii\right)& Abcd& a^3{b}^{3}cd& ab{c}^2{d}^3& -5abcd{p}^2& \end{array}$