Asked by sars
How do you solve 8x^3-24x^2+12x-4=0?
Answers
Answered by
Steve
first step: divide by 4 to make the numbers more manageable:
2x^3 - 6x^2 + 3x - 1 = 0
Now you know that any rational roots will be ±1 or ±1/2
So, try them out. Turns out that none of them is a root.
You can use Descarte's Rule of Signs to show that there are no negative roots, and at most three positive roots (duh).
So, now you have to fall back on the general solution of cubics (very difficult and messy) or iterative methods to get an approximate solution to any desired accuracy.
2x^3 - 6x^2 + 3x - 1 = 0
Now you know that any rational roots will be ±1 or ±1/2
So, try them out. Turns out that none of them is a root.
You can use Descarte's Rule of Signs to show that there are no negative roots, and at most three positive roots (duh).
So, now you have to fall back on the general solution of cubics (very difficult and messy) or iterative methods to get an approximate solution to any desired accuracy.
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