Solving a Cubic Polynomial

[ConnorK]

Hello
I came across a question in my math textbook that i am confused by.

Below is the question.
[imath]2(x-1)^3=32[/imath]

initially i tried collecting everything to one side, expanding all the terms and then factorizing the new cubic polynomial using the factor theorem and rational root theorem, however i found that this did not work when i tried it as none of the possible values values i got from using the rational root theorem made the equation equal 0. I then took the simpler approach of dividing both sides by 2, cube rooting the right side and then add 1 to the right side, this worked.

What i am asking is why did my first approach not work. below is a picture of my first approach. All help appreciated.

 

[blamocur]

I can make no sense of your first approach. Where do values of [imath]\pm 1[/imath] and [imath]\pm 17[/imath] come from?
If [imath](x-1)^3 = 16[/imath] can't you figure out the value of [imath]x-1[/imath] ?

 

[Dr.Peterson]

Below is the question.
[imath]2(x-1)^3=32[/imath]

initially i tried collecting everything to one side, expanding all the terms and then factorizing the new cubic polynomial using the factor theorem and rational root theorem, however i found that this did not work when i tried it as none of the possible values values i got from using the rational root theorem made the equation equal 0. I then took the simpler approach of dividing both sides by 2, cube rooting the right side and then add 1 to the right side, this worked.

What i am asking is why did my first approach not work. below is a picture of my first approach. All help appreciated.

View attachment 38481

The rational root theorem, which you used here, can only find rational roots. Since the solutions are not rational, they are not found this way.

 

[ConnorK]

The rational root theorem, which you used here, can only find rational roots. Since the solutions are not rational, they are not found this way.

Ah that makes sense. Thank you.