Solution
Step 1
A)The domain refers to all values of x for which the function remains defined. Any value of x that makes the function undefined is out of the domain. This can be achieved by factorizing the denominator of the known function to get the values of x that make the function undefined. All other values of x that make the function defined are part of the domain.
Step 2
B)Factorize the denominator
![\begin{gathered} 6x^2+13x-15=0 \\ \text{Factors required are 18 and -5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/luct905p72l8hl2laie0gnd3dbf5u2dg2m.png)
Hence factorizing further we will have
![\begin{gathered} 6x^2+18x\text{ -5x -15=0} \\ (6x^2+18x)(-5x-15)=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/1e3ks35zirvn1l4hthzb7bo3g2ev7n3wwu.png)
![\begin{gathered} 6x(x+3)\text{ -5(x+3) = 0} \\ (6x-5)(x+3)\text{ = 0} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ox4g3s1yq72pun9wvn41v3ej7gt97ws17c.png)
![\begin{gathered} 6x-5\text{ = 0} \\ 6x\text{ =5} \\ x\text{ = }(5)/(6) \\ or \\ x+3\text{ = 0} \\ x\text{ = -3} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vbyyfqtf2zenqhuxjk2zmga5my47khhwnk.png)
Hence, the domain of the function given is all real values of x except -3 and 5/6