Answer:
The third: [tex]\bold{\dfrac{x+5}{x+15}}[/tex]
Step-by-step explanation:
[tex]x^2+19x+70\ \implies a=1\,,\ b=19\,,\ c=70\\\\x=\frac{-19\pm\sqrt{19^2-4\cdot1\cdot70}}{2\cdot1}=\frac{-19\pm\sqrt{361-280}}{2}=\frac{-19\pm9}{2}\ \Rightarrow\ x_1=-14\,,\ x_2=-5\\\\x^2+19x+70=(x+14)(x+5)\\\\\\x^2-225=x^2-(15)^2=(x-15)(x+15)\\\\\\x^2-5x-150\ \implies a=1\,,\ b=-5\,,\ c=-150\\\\x=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot1\cdot(-150)}}{2\cdot1}=\frac{5\pm\sqrt{25+600}}{2}=\frac{5\pm25}{2}\ \Rightarrow\ x_1=-10\,,\ x_2=15\\\\x^2-5x-150=(x+10)(x-15)[/tex]
[tex]x^2+24x+140\ \implies a=1\,,\ b=24\,,\ c=140\\\\x=\frac{-24\pm\sqrt{24^2-4\cdot1\cdot140}}{2\cdot1}=\frac{-24\pm\sqrt{576-560}}{2}=\frac{-24\pm4}{2}\ \Rightarrow\ x_1=-14\,,\ x_2=-10\\\\x^2-5x-150=(x+14)(x+10)[/tex]
[tex]\dfrac{x^2+19x+70}{x^2-225}\,\cdot\,\dfrac{x^2-5x-150}{x^2+24x+140}=\dfrac{(x+14)(x+5)}{(x-15)(x+15)}\cdot\dfrac{(x+10)(x-15)}{(x+14)(x+10)}=\\\\\\=\dfrac{(x+14)(x+5)}{(x-15)(x+15)}\cdot\dfrac{x-15}{x+14}=\dfrac{x+5}{x+15}\cdot\dfrac11=\boxed{\dfrac{x+5}{x+15}}[/tex]
