Question

How do you find the product \[(2a + 9)(5a - 6)\] ?

Answer 1

Hint: Here we are asked to find the product of \[(2a + 9)(5a - 6)\] . So here we will multiply each individual term given in the left parenthesis with the individual term provided in the right parenthesis. With the help of distributive law of multiplication we can find the product by considering \[(2a + 9)\] and \[(5a - 6)\] as two binomials.
The formula used for solving the above question is most commonly seen as a horizontal distributive set up. As there are two binomials so firstly we will multiply (distribute) this term with each of the terms in the second binomial. Later on we'll take the second term in the first binomial and distribute this term times each of the terms in the second binomial. Lastly we will add the terms after combining them.

Complete step by step solution:
So here through the use of distributive law of multiplication we will multiply each individual term by each individual term in the right parenthesis and after combining the like terms we will find the product
 \[
   \Rightarrow (2a + 9)(5a - 6) \\
   \Rightarrow (2a(5a - 6) + 9(5a - 6) \\
   \Rightarrow 10{a^2} - 12a + 45a - 54 \;
 \]
Now we will combine the like terms and get
 \[10{a^2} + 33a - 54\]
So the product of \[(2a + 9)(5a - 6)\] comes out to be \[10{a^2} + 33a - 54\]
So, the correct answer is “ \[10{a^2} + 33a - 54\] ”.

Note: While multiplying two binomials, there are four multiplications done in which each of the first two terms is multiplied with each of the second two terms and the order of multiplication can be in any order.
Notes: The above given equation can be solved by the use of column distributive setup also called as vertical distributive set up. FOIL method is used here which is a technique for distributing two binomials. Where First means to multiply the term occurring in each binomial, Outer means to multiply the outermost term in the product while Inner term stands for multiplying the innermost terms and Last means to multiply the term occurring lastly in each binomial.