# Algebra Bloody Hard is this right

1. Apr 3, 2005

### dagg3r

hey everyone i got a series of questions so can somebody check my working out and see if they are right thanks.

1. Make t the subject of the formula 1/g= 1/(t+1) - 1/(s-1)
i did all the working out and got at the end
t= GS-2G-S+1 / S-1+G not sure if i should make it simplier like
t= G(S-2) - S + 1 / G+s-1

2. solve the equation 6x^3 + 8 = 23x^2 + 6x
i can only get the intercepts using my calculator sketching both graphs and finding the intercepts how do you do it manually????? i tried bringing them all to the LHS and let it equal to zero and subing x to get 0 on the RHS but never happened is there a way i dont think so!!! if there is plz tell me

3. Write in simplest index form:
sqrt ( x^8 * y^4 * z^2) / x*y*z^2

i got totally confused do i just do [(x^8*y^4*z^2)^(1/2) ] / xyz^2
that gives me
= x^4*y^2*z / xyz^2 = x^3*Y*z^-1
therefore
= x^3*Y / Z IS THAT RIGHT PEOPLE???? PLZ HELP THANKS

THATS ALL PLZ HELP ME EVERY1 HEHE :P IT WILL BE MOST APPRECIATED

2. Apr 3, 2005

### Data

1. is right.

For 2., note that $x=4$ satisfies the equation. Then use long division to get

$$6x^3 - 23x^2 - 6x + 8 = (x - 4)(6x^2 + x - 2)$$

so all you need to solve now is

$$6x^2 + x - 2 = 0$$

which should be easy enough...

3. is nearly right. Should be

$$\frac{x^3y}{|z|}$$