(2x^2+8x-42)/(x^2+12+35) Be sure to list restrictions. Can someone do this problem for me? i don't know how to do it and it says show work.

(2*(x+7))*(x-3)/(x^2+47) factor version for you

basicaly it is 3 and -7

what?

i need to show work >.>

(2*(x+7))*(x-3)/(x^2+47)=0 (2*(x+7))*(x-3)=0 multiply by (x^2+47)

ok here it goes step by step

You wanna find the roots? set it equal to zero (2x^2+8x-42)/(x^2+12+35) =0 (2x^2+8x-42)=0 multiply by (x^2+12+35) both sides (2*(x+7))*(x-3)=0 now factor lhs Can you solve from there?

sorry i fail at math i can't

from here it is easy once you factor and quation to () * () = 0 form all you have to do is this (x+7))*(x-3)=0 divide by 2 to get rid of the 2 in lhs now it is in () * ()=0 form so you can solve for each ( ) like this (x+7)=0 and (x-3)=0 subtract 7 add 3 x=-7 x=3 that is all :)

are there restrictions?

well at this point if you wanna find the roots then that is all. if you want the inverse of your function then using the quadratic formula [x = (4+sqrt(100+52*y-47*y^2))/(-2+y)], [x = -(-4+sqrt(100+52*y-47*y^2))/(-2+y)] -2+y cant = 0

thanks

or if you want y=(2x^2+8x-42)/(x^2+12+35) then the denominator cant be 0

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