Question

Find the following product: \[\left( {5x + 7} \right) \times \left( {3x + 4} \right)\]
A) \[15{x^2} - 41x + 14\]
B) \[15{x^3} + 41x + 14\]
C) \[15{x^2} + 41x + 28\]
D) \[15{x^3} + 41x + 28\]

Answer 1

Hint:
Here, we will find the product of the given Binomials. We will multiply the binomials by using the horizontal method or FOIL method to find the product of their Binomials. A binomial expression is defined as an algebraic expression having two terms and these terms must be unlike.

Complete step by step solution:
We are given with an equation \[\left( {5x + 7} \right) \times \left( {3x + 4} \right)\]
We will find the product of the given binomials by FOIL method.
FOIL method is a method of multiplying the binomials by multiplying the first terms, then the outer terms, then the inner terms and at last the last terms. We will find the product of two binomials the result will be a trinomial.
According to this method we will multiply one term of a binomial with each term in the another binomial.
By applying FOIL method, we get
\[\left( {5x + 7} \right) \times \left( {3x + 4} \right) = 5x\left( {3x + 4} \right) + 7\left( {3x + 4} \right)\]
\[ \Rightarrow \left( {5x + 7} \right) \times \left( {3x + 4} \right) = 15{x^2} + 20x + 21x + 28\]
By adding the like terms, we get
\[ \Rightarrow \left( {5x + 7} \right) \times \left( {3x + 4} \right) = 15{x^2} + 41x + 28\]
Therefore, the product of \[\left( {5x + 7} \right) \times \left( {3x + 4} \right)\] is \[15{x^2} + 41x + 28\].

Thus, option (C) is the correct answer.

Note:
We can also find the product by using an algebraic identity.
We are also given with an equation \[\left( {5x + 7} \right) \times \left( {3x + 4} \right)\] which can be written as :
\[\left( {5x + 7} \right) \times \left( {3x + 4} \right) = 5\left( {x + \dfrac{7}{5}} \right) \times 3\left( {x + \dfrac{4}{3}} \right)\]
\[ \Rightarrow \left( {5x + 7} \right) \times \left( {3x + 4} \right) = 15\left( {x + \dfrac{7}{5}} \right) \times \left( {x + \dfrac{4}{3}} \right)\]
The product of Binomials is given by the formula \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\]
By substituting \[a = \dfrac{7}{5}\] and \[b = \dfrac{4}{3}\] in the formula, we get
\[ \Rightarrow 15\left( {x + \dfrac{7}{5}} \right) \times \left( {x + \dfrac{4}{3}} \right) = 15\left[ {{x^2} + \left( {\dfrac{7}{5} + \dfrac{4}{3}} \right)x + \left( {\dfrac{7}{5}} \right)\left( {\dfrac{4}{3}} \right)} \right]\]
By cross multiplying the equation, we get
\[ \Rightarrow 15\left( {x + \dfrac{7}{5}} \right) \times \left( {x + \dfrac{4}{3}} \right) = 15\left[ {{x^2} + \left( {\dfrac{7}{5} \times \dfrac{3}{3} + \dfrac{4}{3} \times \dfrac{5}{5}} \right)x + \left( {\dfrac{7}{5}} \right)\left( {\dfrac{4}{3}} \right)} \right]\]
By simplifying the equation, we get
\[ \Rightarrow 15\left( {x + \dfrac{7}{5}} \right) \times \left( {x + \dfrac{4}{3}} \right) = 15\left[ {{x^2} + \left( {\dfrac{{21 + 20}}{{15}}} \right)x + \left( {\dfrac{{28}}{{15}}} \right)} \right]\]
By multiplying the terms, we get
\[ \Rightarrow 15\left( {x + \dfrac{7}{5}} \right) \times \left( {x + \dfrac{4}{3}} \right) = \left[ {15{x^2} + 15\left( {\dfrac{{21 + 20}}{{15}}} \right)x + 15\left( {\dfrac{{28}}{{15}}} \right)} \right]\]
By cancelling the similar terms, we get
\[ \Rightarrow 15\left( {x + \dfrac{7}{5}} \right) \times \left( {x + \dfrac{4}{3}} \right) = \left[ {15{x^2} + 41x + 28} \right]\]
Thus, we get \[\left( {5x + 7} \right) \times \left( {3x + 4} \right) = 15{x^2} + 41x + 28\].
We should know that when two linear equations of a monomial expression are multiplied, then their product is always a quadratic equation.