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**Hint:**A trinomial is a polynomial which has three terms which are connected by plus or minus symbols. Trinomials are often in the form ${x^2} + bx + c$ but it is not true for all the cases. Trinomials which are in the form of ${x^2} + bx + c$ can be factored as the product of two binomials. Here, in this question we can factor the given trinomial by grouping.

**Complete step by step solution:**

Given a trinomial, $10{x^2} + 11x - 6 = 0$.

In order to solve this given polynomial, we must first look for two numbers which meet the following characteristics:

1. Both the numbers must have a product same as the product of first and last coefficients $\left( {10 \times - 6 = - 60} \right)$.

2. Both the numbers must have a sum same as the middle term $\left( {11} \right)$.

Now, we have to examine all those factor pairs whose product is $ - 60$ and their sum is $11$. Some of the possible factor pairs are:

$ - 1$ and $60$, $1$ and $ - 60$

$ - 2$ and $30$, $2$ and $ - 30$

$ - 3$ and $20$, $3$ and $ - 20$

$ - 4$ and $15$, $4$ and $ - 15$

$ - 5$ and $12$, $5$ and $ - 12$

$ - 6$ and $10$, $6$ and $ - 10$

The only pair whose sum is equal to$11$ is $ - 4,15$.

Since, $15x - 4x = 11x$, so we replace $11x$ in the given trinomial by $15x - 4x$. After replacing we get,

$

\Rightarrow 10{x^2} + 11x - 6 = 0 \\

\Rightarrow 10{x^2} + 15x - 4x - 6 = 0 \\

$

Now, we can factor the obtained polynomial by grouping. We start grouping by sorting the trinomial into groups of two.

$ \Rightarrow \left( {10{x^2} + 15x} \right) - \left( {4x + 6} \right) = 0$

Factor the common term from both the groups.

$ \Rightarrow 5x\left( {2x + 3} \right) - 2\left( {2x + 3} \right) = 0$

Taking the term$\left( {2x + 3} \right)$common, we get,

$ \Rightarrow \left( {2x + 3} \right)\left( {5x - 2} \right) = 0$

Thus, this is trinomial completely factored. This solution can be used to find the values of $x$. The values of $x$ can be calculated by putting the terms $\left( {2x + 3} \right)$ and $\left( {5x - 2} \right)$ equal to zero.

$

\Rightarrow 2x + 3 = 0 \\

\Rightarrow x = - \dfrac{3}{2} \\

$ $

\Rightarrow 5x - 2 = 0 \\

\Rightarrow x = \dfrac{2}{5} \\

$

**Hence, this is our required solution.**

**Note:**Students should remember that since the product of the first and last term is negative, one number of the factor pair should be negative and moreover, this means that the sum of $11$ should be created by the difference of those factor pair’s numbers. To check if values of $x$ are correct or not, students can substitute the desired values of $x$ in the given polynomial and see if LHS=RHS, then their solution is correct.

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