Question

PLEASE help me with this question! I really need help...

Answer 1

The answer is option 3.

Explanation:

First, you have to factorize the expressions :

`\frac{ {x}^(2) + 19x + 70 }{ {x}^(2) - 225 } * \frac{ {x}^(2) - 5x - 150}{ {x}^(2)24x + 140 }`

`= ((x + 5)(x + 14))/((x + 15)(x - 15)) * ((x - 15)(x + 10))/((x + 10)(x + 14))`

Next, you have to cut out the common terms like (x + 14), (x - 15) and (x + 10):

`((x + 5)(x + 14))/((x + 15)(x - 15)) * ((x - 15)(x + 10))/((x + 10)(x + 14))`

`= (x + 5)/(x + 15)`

Answer 2

The third:
`\bold{(x+5)/(x+15)}`

Explanation:

`x^2+19x+70\ \implies a=1\,,\ b=19\,,\ c=70\\\\x=(-19\pm√(19^2-4\cdot1\cdot70))/(2\cdot1)=(-19\pm√(361-280))/(2)=(-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=(-(-5)\pm√((-5)^2-4\cdot1\cdot(-150)))/(2\cdot1)=(5\pm√(25+600))/(2)=(5\pm25)/(2)\ \Rightarrow\ x_1=-10\,,\ x_2=15\\\\x^2-5x-150=(x+10)(x-15)`

`x^2+24x+140\ \implies a=1\,,\ b=24\,,\ c=140\\\\x=(-24\pm√(24^2-4\cdot1\cdot140))/(2\cdot1)=(-24\pm√(576-560))/(2)=(-24\pm4)/(2)\ \Rightarrow\ x_1=-14\,,\ x_2=-10\\\\x^2-5x-150=(x+14)(x+10)`

`(x^2+19x+70)/(x^2-225)\,\cdot\,(x^2-5x-150)/(x^2+24x+140)=((x+14)(x+5))/((x-15)(x+15))\cdot((x+10)(x-15))/((x+14)(x+10))=\\\\\\=((x+14)(x+5))/((x-15)(x+15))\cdot(x-15)/(x+14)=(x+5)/(x+15)\cdot\frac11=\boxed{(x+5)/(x+15)}`