How do you expand and simplify \[(2x - 3)(3x + 5)\]?

Hint: Here we have to expand and simplify the given expression. Now, we will multiply any term by another one, as the associative law of multiplication is an identity property. On doing some simplification we get the required answer.

Complete Step by Step Solution:
We have to find the product of \[(2x - 3)\] and \[(3x + 5)\].
So, we can write this in the following way:
\[ \Rightarrow (2x - 3) \times (3x + 5)\]
So, firstly we will multiply the first expression by the first term in the second expression and then we will multiply the first expression by the second term in the second expression.
Now, we need to multiply the first terms and second terms accordingly:
\[ \Rightarrow (2x - 3) \times (3x + 5)\]
\[ \Rightarrow \{ (2x - 3) \times 3x\} + \{ (2x - 3) \times 5\} .\]
Now, perform the rest of the multiplication, we get:
\[ \Rightarrow \{ (2 \times 3 \times {x^2}) - (3 \times 3x)\} + \{ (2x \times 5) - (3 \times 5)\} .\]
By performing further multiplication and addition, we get:
\[ \Rightarrow (6{x^2} - 9x) + (10x - 15).\]
After re arrangements, we get:
\[ \Rightarrow (6{x^2} - 9x + 10x - 15).\]
Now, we will add the terms that have the variable of the same degree.
So,\[(2x - 3) \times (3x + 5)\]
\[ \Rightarrow (6{x^2} + x - 15).\]

\[\therefore \]The answer of the product is \[(6{x^2} + x - 15).\]

Note: Points to remember:
We need to add those variable terms that have the same degree.
Also, if constant terms are more than one, we need to add them separately.
Algebraic production of two terms defines by the following rules:
\[(1)\] Multiply only same degree variables.
 \[(2)\] Multiplication of ‘\[ + \]’ and ‘\[ - \]’ gives us the ‘\[ - \]’ sign always.
\[(3)\] Always add same degree variables with variables and constant terms with constant terms.
\[(4)\]BODMAS rule is the same as the rest of multiplications.
Another way of solution:
Put the same variables and signs at the above formula:
\[(2x - 3) \times (3x + 5) = (6{x^2} - 9x) + (10x - 15)\]
\[ = (6{x^2} + x - 15)\]