Question

(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.

Answer 1

(2*(x+7))*(x-3)/(x^2+47)

factor version for you

Answer 2

basicaly it is 3 and -7

Answer 3

what?

Answer 4

i need to show work >.>

Answer 5

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

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

Answer 6

ok here it goes step by step

Answer 7

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?

Answer 8

sorry i fail at math i can't

Answer 9

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 :)

Answer 10

are there restrictions?

Answer 11

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

Answer 12

thanks

Answer 13

or if you want y=(2x^2+8x-42)/(x^2+12+35)

then the denominator cant be 0