Question & Answer
QUESTION

Use the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to find the product $\left( 2a+4b \right)\left( 2a-4b \right)$.

ANSWER Verified Verified
Hint: Let A = 2a and B = 4b. Use the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Revert to original variables and use the law of exponents to simplify the expression.

Complete step-by-step solution -

Finding the product of two algebraic expressions:
When we multiply two algebraic expressions, each term of one expression is multiplied with all the terms of the expressions. The sum is then simplified, and the result is the product of the two algebraic expressions.
Consider the algebraic expression a+b and other algebraic expression a-b.
When we multiply the two expressions, we multiply each term of a+b to all the terms of a-b.
Now a(a) $={{a}^{2}}$
a(-b) = -ab
b(a) = ba = ab.
b(-b) $=-{{b}^{2}}$
Hence $\left( a+b \right)\left( a-b \right)={{a}^{2}}-ab+ba-{{b}^{2}}$
Now, the terms ab and -ba cancel each other.
So, we have $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
We will use this identity to solve the above question.
Let A = 2a and B = 4b.
So we have (2a+4b)(2a-4b) =(A+B)(A-B)
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
Hence, we have
$\left( A+B \right)\left( A-B \right)={{A}^{2}}-{{B}^{2}}$.
Reverting to original variables, we have
$\left( 2a+4b \right)\left( 2a-4b \right)={{\left( 2a \right)}^{2}}-{{\left( 4b \right)}^{2}}$
Now, we know that ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$.
Hence, we have
${{\left( 2a \right)}^{2}}={{2}^{2}}{{a}^{2}}=4{{a}^{2}}$ and ${{\left( 4b \right)}^{2}}={{4}^{2}}{{b}^{2}}=16{{b}^{2}}$
Hence, we have $\left( 2a+4b \right)\left( 2a-4b \right)=4{{a}^{2}}-16{{b}^{2}}$.

Note: [1] Algebraic identities to memorise.
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
 & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \\
\end{align}$
[2] We can find the product without the use of identities also
We have $2a\left( 2a \right)=4{{a}^{2}}$, $\left( 2a \right)\left( -4b \right)=-8ab$, $\left( 4b \right)\left( 2a \right)=8ab$ and $\left( 4b \right)\left( -4b \right)=-16{{b}^{2}}$
Hence $\left( 2a+4b \right)\left( 2a-4b \right)=4{{a}^{2}}-8ab+8ab-16{{b}^{2}}$
The terms 8ab and -8ab cancel each other.
So, we have
$\left( 2a+4b \right)\left( 2a-4b \right)=4{{a}^{2}}-16ab$, which is the same as obtained above.
The use of identities makes the simplification of algebraic expressions easy.