Answer

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**Hint:**

Here, we have to simplify the given Binomials. We will multiply the binomials by using the horizontal method or FOIL method to find the product of their Binomials. 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. Using this It is possible to combine like terms.

**Complete step by step solution:**

We are given with an equation \[\left( {2a - 1} \right)\left( {3a + 2} \right)\]

We will find the product of the given binomials by using the horizontal method or the FOIL method of multiplying one term of a binomial with each other term in another binomial.

By multiplying each term in one binomial with each other term in the other binomial, we get

\[ \Rightarrow \left( {2a - 1} \right)\left( {3a + 2} \right) = 2a\left( {3a + 2} \right) - 1\left( {3a + 2} \right)\]

Again multiplying the terms, we get

\[ \Rightarrow \left( {2a - 1} \right)\left( {3a + 2} \right) = 6{a^2} + 4a - 3a - 2\]

By subtracting the like terms, we get

\[ \Rightarrow \left( {2a - 1} \right)\left( {3a + 2} \right) = 6{a^2} + a - 2\]

**Therefore, the product of two binomials \[\left( {2a - 1} \right)\left( {3a + 2} \right)\] is \[6{a^2} + a - 2\]**

**Note:**

We know that a binomial expression is defined as an algebraic expression having two terms and these terms must be unlike. We should know that when two linear equation of a monomial expression are multiplied, then their product is always a quadratic equation. We can also find the product by using an algebraic identity.

The given equation \[\left( {2a - 1} \right)\left( {3a + 2} \right)\] can also be written as

\[\left( {2a - 1} \right)\left( {3a + 2} \right) = 2\left( {a - \dfrac{1}{2}} \right) \times 3\left( {a + \dfrac{2}{3}} \right)\]

Multiplying the terms, we get

\[ \Rightarrow \left( {2a - 1} \right) \times \left( {3a + 2} \right) = 6\left( {a - \dfrac{1}{2}} \right) \times \left( {a + \dfrac{2}{3}} \right)\]

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 \[a = - \dfrac{1}{2}\] and \[b = \dfrac{2}{3}\] in the above formula, we get

\[ \Rightarrow 6\left( {a - \dfrac{1}{2}} \right) \times \left( {a + \dfrac{2}{3}} \right) = 6\left[ {{a^2} + \left( { - \dfrac{1}{2} + \dfrac{2}{3}} \right)a + \left( { - \dfrac{1}{2}} \right)\left( {\dfrac{2}{3}} \right)} \right]\]

Cross multiplying the equation, we get

\[ \Rightarrow 6\left( {a - \dfrac{1}{2}} \right) \times \left( {a + \dfrac{2}{3}} \right) = 6\left[ {{a^2} + \left( { - \dfrac{1}{2} \times \dfrac{3}{3} + \dfrac{2}{3} \times \dfrac{2}{2}} \right)a + \left( { - \dfrac{1}{2}} \right)\left( {\dfrac{2}{3}} \right)} \right]\]

By simplifying the equation, we get

\[ \Rightarrow 6\left( {a - \dfrac{1}{2}} \right) \times \left( {a + \dfrac{2}{3}} \right) = 6\left[ {{a^2} + \left( {\dfrac{{ - 3 + 4}}{6}} \right)a + \left( { - \dfrac{1}{3}} \right)} \right]\]

Multiplying the terms, we get

\[ \Rightarrow 6\left( {a - \dfrac{1}{2}} \right) \times \left( {a + \dfrac{2}{3}} \right) = \left[ {6{a^2} + 6\left( {\dfrac{1}{6}} \right)a + 6\left( { - \dfrac{1}{3}} \right)} \right]\]

By cancelling the similar terms, we get

\[ \Rightarrow 6\left( {a - \dfrac{1}{2}} \right) \times \left( {a + \dfrac{2}{3}} \right) = \left[ {6{a^2} + a - 2} \right]\]

Therefore, the product of two binomials \[\left( {2a - 1} \right)\left( {3a + 2} \right)\] is\[6{a^2} + a - 2\] .

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