Answer:
[tex]\frac{y+5}{y-3} \textrm{ where } y \ne 8[/tex]
Step-by-step explanation:
We'll first factorize the quadratic equations by finding their zero-points:
[tex]y^2 - 3y - 40 = 0\\\Delta = (-3)^2 - 4\cdot 1 \cdot -40 = 9 + 160 = 169\\\sqrt{\Delta} = 13\\y = \frac{3 \pm 13}{2}\\y_1 = 8\\y_2 = -5\\y^2 - 3y - 40 = (y - 8)(y + 5)\\~\\y^2 - 11y + 24 = 0\\\Delta = (-11)^2 - 4\cdot 1\cdot 24 = 121 - 96 = 25\\\sqrt{\Delta}} = 5\\y = \frac{11 \pm 5}{2}\\y_1 = 8\\y_2 = 3\\y^2 - 11y + 24 = (y - 8)(y - 3)[/tex]
Now we can use this to simplify the whole division:
[tex]\frac{y^2-3y-40}{y^2-11y+24} = \frac{(y-8)(y+5)}{(y-8)(y-3)} = \frac{y+5}{y-3} \textrm{ where } y \ne 8[/tex]
