Find all roots to the cubic equation x^3 - 5x^2 + 8x + 4 (no calculator)

Question

precal · Answer

3 solutions because of the third power

precal · Answer

1 real and 2 complex|dw:1344796330060:dw|

anonymous · Answer

without a calculator, good luck

fwizbang · Answer

Since the coefficients are integers, try guessing small integers (....1,0,-1....). If you find an answer a that works, divide x-a into the polynomial to turn it into a quadratic. Then factor the quadratic to find the others.

anonymous · Answer

except that it does not work in this case

anonymous · Answer

only possible rational zeros are $\pm1,\pm2$ and it is none of these

anonymous · Answer

it looks like 1 and 2 work.

anonymous · Answer

but when I'm dividing the equation by x-a (both 2 and 1 worked just plugging them in so I picked 2) It only worked when I was dividing it positively like x+ 2 instead of x-2. Am I doing it wrong?

anonymous · Answer

I got x=2 (multiplicity of 2) and x=1 , but the negative and positive is so screwy, it worked with dividing by x+2 but not with x-2 and plugging in 2 and 1 worked. Trying to divided by x-a instead of x+a didn't work.

fwizbang · Answer

I missed the - sign on the 5 on the sidebar, sorry. sat73 is right. Are you sure the signs are all listed correctly? If so the general procedure is listed here http://mathworld.wolfram.com/CubicFormula.html I'd be sure that all the - signs on the coefficients are correct before I did this though.

anonymous · Answer

well it all equals zero if that makes a difference, but the signs are all right. (my math teacher has been known to occasionaly put the wrong signs)

fwizbang · Answer

Are the roots supposed to be real?

anonymous · Answer

I believe so, this was prep summer assignment for my AP calc class, so there's got to be a simple way to do this.

precal · Answer

http://www.wolframalpha.com/input/?i=x%5E3+-+5x%5E2+%2B+8x+%2B+4%3D0 only one real solution, not an easy one to do by hand. If it is for an AP calc class, it is possible that you were to do it with a calculator. This would be a good problem that requires the use of a calculator. AP won't give you something this ugly to do by hand for a non calculator problem.

anonymous · Answer

This was the part of the summer assignment given to me by my AP calc teacher, and it said no calculator next to it.

fwizbang · Answer

The roots of the cubic as written, are complex, so there's probably a sign error in the problem. If you change the -5 to +5, or the 4 to -4, I think its doable.