
Find the product of:
\[\begin{align}
& (i)\left( {{p}^{2}}-{{q}^{2}} \right)\left( 2p+q \right) \\
& (ii)\left( x+3y \right)\left( x-y \right) \\
\end{align}\]
Answer
596.7k+ views
Hint: Here, in the given question, we can proceed the question by using formula or an identity that, if there are four expressions a, b, c, d then the product of (a+b) and (c+d) will be ac + ad + bc + bd to find the desired results or solutions.
Complete step-by-step answer:
In the question, we are given two parts and in first part we have to find products of two expressions which are \[\left( {{p}^{2}}-{{q}^{2}} \right)\text{ and }\left( 2p+q \right)\] and in the second part we have to find product of two expressions which are \[\left( x+3y \right)\text{ and }\left( x-y \right)\]
(i) So, the two expression are \[\left( {{p}^{2}}-{{q}^{2}} \right)\text{ and }\left( 2p+q \right)\] so their product will be calculated by using the following identity or formula for the terms a, b, c, d the product of (a+b) (c+d) will be equal to ac + ad + bc + bd.
So, here it is given \[\left( {{p}^{2}}-{{q}^{2}} \right)\left( 2p+q \right)\] so we can use the above identity by considering \[a\text{ as }{{\text{p}}^{\text{2}}},\text{ b as -}{{\text{q}}^{\text{2}}},\text{ c as 2p and d as q}\]
Thus, according to the identity the product equals to \[\left( {{p}^{2}} \right)\times \left( 2p \right)+\left( {{p}^{2}} \right)\times \left( q \right)+\left( -{{q}^{2}} \right)\times \left( 2p \right)+\left( -{{q}^{2}} \right)\times \left( q \right)\]
On further calculation it equals to \[2{{p}^{3}}+{{p}^{2}}q-2p{{q}^{2}}-{{q}^{3}}\]
Hence, the product of \[\left( {{p}^{2}}-{{q}^{2}} \right)\left( 2p+q \right)\] is \[2{{p}^{3}}+{{p}^{2}}q-2p{{q}^{2}}-{{q}^{3}}\]
(ii) So, the two expression are \[\left( x+3y \right)\text{ and }\left( x-y \right)\] so their product will be calculated by using the following identity or formula for the term a, b, c, d the product of (a+b) (c+d) will be equal to ac + ad + bc + bd.
So, here it is given \[\left( x+3y \right)\left( x-y \right)\] so, we can use the above identity by considering \[\text{a as x, b as 3y, c as x and d as -y}\]
Thus, according to the identity the product equals to \[\left( x \right)\times \left( x \right)+\left( x \right)\times \left( -y \right)+\left( 3y \right)\times \left( x \right)+\left( 3y \right)\times \left( -y \right)\]
On further calculation it equals to \[{{x}^{2}}-xy+3xy-3{{y}^{2}}\]
Hence, the product of \[\left( x+3y \right)\left( x-y \right)\] is ${{x}^{2}}-xy+3xy-3{{y}^{2}}$.
Note: We can also check that whether the products one gets is correct or not by factoring them and if on factoring if it comes to the original question then, the answer is correct otherwise wrong. In the first expression, the terms are \[\left( {{p}^{2}}-{{q}^{2}} \right)\text{ and }\left( 2p+q \right)\] , we can also write the first term as (p+q)(p-q) and then expand by multiplying the three terms. But, this will simply increase the calculation further, so this simplification of first terms is not useful here.
Complete step-by-step answer:
In the question, we are given two parts and in first part we have to find products of two expressions which are \[\left( {{p}^{2}}-{{q}^{2}} \right)\text{ and }\left( 2p+q \right)\] and in the second part we have to find product of two expressions which are \[\left( x+3y \right)\text{ and }\left( x-y \right)\]
(i) So, the two expression are \[\left( {{p}^{2}}-{{q}^{2}} \right)\text{ and }\left( 2p+q \right)\] so their product will be calculated by using the following identity or formula for the terms a, b, c, d the product of (a+b) (c+d) will be equal to ac + ad + bc + bd.
So, here it is given \[\left( {{p}^{2}}-{{q}^{2}} \right)\left( 2p+q \right)\] so we can use the above identity by considering \[a\text{ as }{{\text{p}}^{\text{2}}},\text{ b as -}{{\text{q}}^{\text{2}}},\text{ c as 2p and d as q}\]
Thus, according to the identity the product equals to \[\left( {{p}^{2}} \right)\times \left( 2p \right)+\left( {{p}^{2}} \right)\times \left( q \right)+\left( -{{q}^{2}} \right)\times \left( 2p \right)+\left( -{{q}^{2}} \right)\times \left( q \right)\]
On further calculation it equals to \[2{{p}^{3}}+{{p}^{2}}q-2p{{q}^{2}}-{{q}^{3}}\]
Hence, the product of \[\left( {{p}^{2}}-{{q}^{2}} \right)\left( 2p+q \right)\] is \[2{{p}^{3}}+{{p}^{2}}q-2p{{q}^{2}}-{{q}^{3}}\]
(ii) So, the two expression are \[\left( x+3y \right)\text{ and }\left( x-y \right)\] so their product will be calculated by using the following identity or formula for the term a, b, c, d the product of (a+b) (c+d) will be equal to ac + ad + bc + bd.
So, here it is given \[\left( x+3y \right)\left( x-y \right)\] so, we can use the above identity by considering \[\text{a as x, b as 3y, c as x and d as -y}\]
Thus, according to the identity the product equals to \[\left( x \right)\times \left( x \right)+\left( x \right)\times \left( -y \right)+\left( 3y \right)\times \left( x \right)+\left( 3y \right)\times \left( -y \right)\]
On further calculation it equals to \[{{x}^{2}}-xy+3xy-3{{y}^{2}}\]
Hence, the product of \[\left( x+3y \right)\left( x-y \right)\] is ${{x}^{2}}-xy+3xy-3{{y}^{2}}$.
Note: We can also check that whether the products one gets is correct or not by factoring them and if on factoring if it comes to the original question then, the answer is correct otherwise wrong. In the first expression, the terms are \[\left( {{p}^{2}}-{{q}^{2}} \right)\text{ and }\left( 2p+q \right)\] , we can also write the first term as (p+q)(p-q) and then expand by multiplying the three terms. But, this will simply increase the calculation further, so this simplification of first terms is not useful here.
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