# To solve the algebraic expression $({{x}^{2}}-{{y}^{2}})\times (x+2y)$.

Last updated date: 22nd Mar 2023

•

Total views: 305.1k

•

Views today: 2.87k

Answer

Verified

305.1k+ views

Hint: While solving this problem it is important to know the multiplication of algebraic expressions. That is, ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$ (where, x is a algebraic variable, a and b are any integers).

Complete step-by-step solution:

To solve any algebraic expression involving basic arithmetic operations like addition, subtraction, multiplication and division, we follow the basic procedures as below-

For addition and subtraction, ax + bx + cy – dy = (a+b) x + (c-d) y

We can combine constants (in this case a and b for x and c and d for y) for the same variable (as shown above).

In case of multiplication and division,

(ax+by)$\times $(cx+dy)

First multiplying ax with (cx+dy) and then by with (cx+dy), we get,

= [(ax)$\times $(cx+dy)] + [(by)$\times $(cx+dy)]

=[ac${{x}^{2}}$+adxy] + [bcxy + bd${{y}^{2}}$]

Now, simply following rules of addition, we get,

= ac${{x}^{2}}$+ (ad+bc)xy + bd${{y}^{2}}$

Now, applying these rules to solve the question given,

=$({{x}^{2}}-{{y}^{2}})\times (x+2y)$

Now multiplying ${{x}^{2}}$ with (x+2y) and then $(-{{y}^{2}})$ with (x+2y), we get,

= [${{x}^{2}}$$\times $ (x+2y)] + [$(-{{y}^{2}})$$\times $(x+2y)]

Now, using the property, ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$, we get,

=[${{x}^{2+1}}+2{{x}^{2}}y$] + [$-{{y}^{2}}x-2{{y}^{2+1}}$]

=\[\]$[{{x}^{3}}+2{{x}^{2}}y]+[-{{y}^{2}}x-2{{y}^{3}}]$

=${{x}^{3}}+2{{x}^{2}}y-{{y}^{2}}x-2{{y}^{3}}$

Hence, the solution to the algebraic expression of the question is ${{x}^{3}}+2{{x}^{2}}y-{{y}^{2}}x-2{{y}^{3}}$.

Note: Alternative way to solve the question is by multiplying (x+2y) by $({{x}^{2}}-{{y}^{2}})$. That is,

(x+2y)$\times $ $({{x}^{2}}-{{y}^{2}})$

That is this time we multiply x by $({{x}^{2}}-{{y}^{2}})$ and then add that by the expression we get by multiplying (2y) by $({{x}^{2}}-{{y}^{2}})$, thus,

[(x)$\times $ $({{x}^{2}}-{{y}^{2}})$]+[2y$\times $$({{x}^{2}}-{{y}^{2}})$]

$\begin{align}

& =[{{x}^{3}}-x{{y}^{2}}]+[2y{{x}^{2}}-2{{y}^{3}}] \\

& ={{x}^{3}}-x{{y}^{2}}+2y{{x}^{2}}-2{{y}^{3}} \\

\end{align}$

=${{x}^{3}}+2{{x}^{2}}y-{{y}^{2}}x-2{{y}^{3}}$

Which is the same expression we got in the solution.

Complete step-by-step solution:

To solve any algebraic expression involving basic arithmetic operations like addition, subtraction, multiplication and division, we follow the basic procedures as below-

For addition and subtraction, ax + bx + cy – dy = (a+b) x + (c-d) y

We can combine constants (in this case a and b for x and c and d for y) for the same variable (as shown above).

In case of multiplication and division,

(ax+by)$\times $(cx+dy)

First multiplying ax with (cx+dy) and then by with (cx+dy), we get,

= [(ax)$\times $(cx+dy)] + [(by)$\times $(cx+dy)]

=[ac${{x}^{2}}$+adxy] + [bcxy + bd${{y}^{2}}$]

Now, simply following rules of addition, we get,

= ac${{x}^{2}}$+ (ad+bc)xy + bd${{y}^{2}}$

Now, applying these rules to solve the question given,

=$({{x}^{2}}-{{y}^{2}})\times (x+2y)$

Now multiplying ${{x}^{2}}$ with (x+2y) and then $(-{{y}^{2}})$ with (x+2y), we get,

= [${{x}^{2}}$$\times $ (x+2y)] + [$(-{{y}^{2}})$$\times $(x+2y)]

Now, using the property, ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$, we get,

=[${{x}^{2+1}}+2{{x}^{2}}y$] + [$-{{y}^{2}}x-2{{y}^{2+1}}$]

=\[\]$[{{x}^{3}}+2{{x}^{2}}y]+[-{{y}^{2}}x-2{{y}^{3}}]$

=${{x}^{3}}+2{{x}^{2}}y-{{y}^{2}}x-2{{y}^{3}}$

Hence, the solution to the algebraic expression of the question is ${{x}^{3}}+2{{x}^{2}}y-{{y}^{2}}x-2{{y}^{3}}$.

Note: Alternative way to solve the question is by multiplying (x+2y) by $({{x}^{2}}-{{y}^{2}})$. That is,

(x+2y)$\times $ $({{x}^{2}}-{{y}^{2}})$

That is this time we multiply x by $({{x}^{2}}-{{y}^{2}})$ and then add that by the expression we get by multiplying (2y) by $({{x}^{2}}-{{y}^{2}})$, thus,

[(x)$\times $ $({{x}^{2}}-{{y}^{2}})$]+[2y$\times $$({{x}^{2}}-{{y}^{2}})$]

$\begin{align}

& =[{{x}^{3}}-x{{y}^{2}}]+[2y{{x}^{2}}-2{{y}^{3}}] \\

& ={{x}^{3}}-x{{y}^{2}}+2y{{x}^{2}}-2{{y}^{3}} \\

\end{align}$

=${{x}^{3}}+2{{x}^{2}}y-{{y}^{2}}x-2{{y}^{3}}$

Which is the same expression we got in the solution.

Recently Updated Pages

If abc are pthqth and rth terms of a GP then left fraccb class 11 maths JEE_Main

If the pthqth and rth term of a GP are abc respectively class 11 maths JEE_Main

If abcdare any four consecutive coefficients of any class 11 maths JEE_Main

If A1A2 are the two AMs between two numbers a and b class 11 maths JEE_Main

If pthqthrth and sth terms of an AP be in GP then p class 11 maths JEE_Main

One root of the equation cos x x + frac12 0 lies in class 11 maths JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?