Hi Tom,

Have a look at the the equation I used to respond to Janie's question,

\[\frac{x+1}{x-2} = \frac{3}{x-2} + 5.\]

Multiplying each side by $(x - 2)$ results in a linear equation with solution $x = 2.$ But $x = 2$ can not be a solution to the original equation as it makes the denominators equal to zero.

You can use something similar with the denominators $(x + 3)$ so that $x = -3$ cannot be a solution, something like

\[\frac{x+1}{x+3} = \frac{a}{x+3} + b.\]

In this case multiplying both sides by $(x + 3)$ yields a linear equation but you want a quadratic equation with $x = -2$ and $x = -3$ as solutions. Try

\[\frac{x^2+1}{x+3} = \frac{a}{x+3} + b.\]

Multiply both sides by $(x + 3)$, simplify and see if you can choose $a$ and $b$ so that the resulting quadratic equation has solutions $x = -2$ and $x = -3.$

Harley