
Use the identity $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$ to find the given product: $\left( 2x+5y \right)\left( 2x+3y \right)$
A. $4{{x}^{2}}+24xy-15{{y}^{2}}$
B. $4{{x}^{2}}+16xy+25{{y}^{2}}$
C. $4{{x}^{2}}+16xy+15{{y}^{2}}$
D. ${{x}^{2}}+16xy+15{{y}^{2}}$
Answer
460.2k+ views
Hint: To obtain the product of the given value we will use the given identity. Firstly we will compare the given identity with the expression whose product we have to find and get the value of $a,b,x$. Then we will equate the obtained value in the identity and solve it to get our desired answer.
Complete step by step answer:
The identity given to us is as follows:
$\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$……$\left( 1 \right)$
We have to find the product of the given expression:
$\left( 2x+5y \right)\left( 2x+3y \right)$……$\left( 2 \right)$
On comparing equation (1) and (2) we get the values as:
$\begin{align}
& x=2x \\
& a=5y \\
& b=3y \\
\end{align}$
Now put the above value in equation (1) and simplify it to get the product as follows:
$\begin{align}
& \left( 2x+5y \right)\left( 2x+3y \right)={{\left( 2x \right)}^{2}}+\left( 5y+3y \right)\times 2x+\left( 5y\times 3y \right) \\
& \Rightarrow \left( 2x+5y \right)\left( 2x+3y \right)=4{{x}^{2}}+8y\times 2x+15{{y}^{2}} \\
& \therefore \left( 2x+5y \right)\left( 2x+3y \right)=4{{x}^{2}}+16xy+15{{y}^{2}} \\
\end{align}$
So we got the product of $\left( 2x+5y \right)\left( 2x+3y \right)$ as $4{{x}^{2}}+16xy+15{{y}^{2}}$
So, the correct answer is “Option C”.
Note: The given identity is known as a special binomial product where the first term of both the sum basic binomials is the same and therefore it is a special case in algebraic mathematics. When we expand the special binomial product we get an algebraic expression. Algebraic identity is a standard formula that can be used for any value of variable in it. It is used to solve factorization of polynomials and the algebraic equations. There are many algebraic identities used in mathematics. Algebraic expressions are the expressions that are formed by variables and constant along with the various algebraic operations such as addition, subtraction, multiplication and division. It is different from an algebraic equation as it doesn’t have any equal sign in it.
Complete step by step answer:
The identity given to us is as follows:
$\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$……$\left( 1 \right)$
We have to find the product of the given expression:
$\left( 2x+5y \right)\left( 2x+3y \right)$……$\left( 2 \right)$
On comparing equation (1) and (2) we get the values as:
$\begin{align}
& x=2x \\
& a=5y \\
& b=3y \\
\end{align}$
Now put the above value in equation (1) and simplify it to get the product as follows:
$\begin{align}
& \left( 2x+5y \right)\left( 2x+3y \right)={{\left( 2x \right)}^{2}}+\left( 5y+3y \right)\times 2x+\left( 5y\times 3y \right) \\
& \Rightarrow \left( 2x+5y \right)\left( 2x+3y \right)=4{{x}^{2}}+8y\times 2x+15{{y}^{2}} \\
& \therefore \left( 2x+5y \right)\left( 2x+3y \right)=4{{x}^{2}}+16xy+15{{y}^{2}} \\
\end{align}$
So we got the product of $\left( 2x+5y \right)\left( 2x+3y \right)$ as $4{{x}^{2}}+16xy+15{{y}^{2}}$
So, the correct answer is “Option C”.
Note: The given identity is known as a special binomial product where the first term of both the sum basic binomials is the same and therefore it is a special case in algebraic mathematics. When we expand the special binomial product we get an algebraic expression. Algebraic identity is a standard formula that can be used for any value of variable in it. It is used to solve factorization of polynomials and the algebraic equations. There are many algebraic identities used in mathematics. Algebraic expressions are the expressions that are formed by variables and constant along with the various algebraic operations such as addition, subtraction, multiplication and division. It is different from an algebraic equation as it doesn’t have any equal sign in it.
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