Question

Perform the indicated operations and simplify.

$\left(2x+y-3\right)\left(2x+y+3\right)$

Short answer

The solution$\left(2x+y-3\right)\left(2x+y+3\right)=4x^2+4xy+y^2-9$

Step-by-step solution

Step 1. Apply the special product formula of “product of the sum and difference of terms”.

If A and B are any real numbers or algebraic expressions, then

$\left(A+B\right)\left(A-B\right)=\left(A-B\right)\left(A+B\right)=A^2-B^2$

Step 2. Apply the square of a sum formula.

If A and B are any real numbers or algebraic expressions, then

${\left(A+B\right)}^2=A^2+2AB+B^2$

Step 3. Apply the concept of addition and subtraction of polynomials.

We add and subtract polynomials using the properties of real numbers. The idea is to combine like terms (that is, terms with the same variables raised to the same powers) using the Distributive Property.

In subtracting polynomials, we have to remember that if a minus sign precedes an expression in parentheses, then the sign of every term within the parentheses is changed when we remove the parentheses.

Step 4. Analyze the given expression.

Given$\left(2x+y-3\right)\left(2x+y+3\right)$

Using product of the sum and difference of terms, we can write it as:

Here$A=2x+y,B=3$

$\left(2x+y-3\right)\left(2x+y+3\right)={\left(2x+y\right)}^2-3^2$

Step 5. Simplify.

Simplify the expression in Step 2:

$\begin{array}{l}\left(2x+y-3\right)\left(2x+y+3\right)={\left(2x+y\right)}^2-3^2\\ \Rightarrow \left(2x+y-3\right)\left(2x+y+3\right)={\left(2x\right)}^2+4xy+y^2-3^2\mathrm{....................}\left[{\left(A+B\right)}^2=A^2+2AB+B^2\right]\\ \Rightarrow \left(2x+y-3\right)\left(2x+y+3\right)=4x^2+4xy+y^2-9\end{array}$