Dividing by a fraction is equivalent to multiply by its reciprocal, then:

Now, we need to express the quadratic polynomials using their roots, as follows:

where y1 and y2 are the roots.
Applying the quadratic formula to the first polynomial:
![\begin{gathered} y_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_(1,2)=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_(1,2)=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=(7+11)/(6)=3 \\ y_2=(7-11)/(6)=-(2)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f6r4e7zgnu31bdummretsa79tg6fy8y5eu.png)
Applying the quadratic formula to the second polynomial:
![\begin{gathered} y_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_(1,2)=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_(1,2)=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=(-1+5)/(4)=1 \\ y_2=(-1-5)/(4)=-(3)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xt1plmwtk8llcf8whi2luybjn33tl4ia0m.png)
Applying the quadratic formula to the third polynomial:
![\begin{gathered} y_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_(1,2)=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_(1,2)=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=(3+9)/(4)=3 \\ y_2=(3-9)/(4)=-(3)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/15i37pptq8oo6bfrsl1znzz4ocz6sh377o.png)
Applying the quadratic formula to the fourth polynomial:
![\begin{gathered} y_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_(1,2)=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_(1,2)=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=(-1+3)/(2)=1 \\ y_2=(-1-3)/(2)=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jojqoweb1vep7fjn1qkerh1sj4nfiz9e11.png)
Substituting into the rational expression and simplifying:
