Thread: Simple Rearanging to solve for x

1. Simple Rearanging to solve for x

Okay so I have to solve for x

$\displaystyle 120 = 2(x-2)(x-3)(x+5)$
$\displaystyle 60 = (x-2)(x-3)(x+5)$
$\displaystyle 60 = x^3 - 19x + 30$

I am stuck from here. I don't know if I should have expanded or not.

2. Originally Posted by JimmyRP
Okay so I have to solve for x

$\displaystyle 120 = 2(x-2)(x-3)(x+5)$
$\displaystyle 60 = (x-2)(x-3)(x+5)$
$\displaystyle 60 = x^3 - 19x + 30$

I am stuck from here. I don't know if I should have expanded or not.
$\displaystyle x^3 - 19x - 30 = 0$

rational root theorem ... try $\displaystyle x = -2$ as your first root, do some synthetic division, and solve for the other two roots.

3. Originally Posted by JimmyRP
Okay so I have to solve for x

$\displaystyle 120 = 2(x-2)(x-3)(x+5)$
$\displaystyle 60 = (x-2)(x-3)(x+5)$
$\displaystyle 60 = x^3 - 19x + 30$

I am stuck from here. I don't know if I should have expanded or not.

now subtract 60 from both sides

$\displaystyle x^3-19x-30=0$

Now we need to use the rational roots theorem. So we know the only roots are factors of 30.

So if we try x=5 we get

$\displaystyle 5^3-19(5)-30=125-95-30=0$

so 5 is a root so x-5 is a factor. So by long division we get

$\displaystyle (x-5)(x^2+5x+6)=0$

$\displaystyle (x-5)(x+2)(x+3)=0$

I hope this helps

4. thanks a lot. I just realized what I had to do, then I checked here and you guys helped. Thanks a lot