Find all the real solutions for the polynomial
equation
x^4 − 5x^3 − 30x^2 − 30x − 216 = 0.
1. x = −4, 9,−1
2. x = −4, 9
3. x = ±1, 6, 10
4. x = 1,−6, 10
5. x = −10, 6
6. None of these

Question

Find all the real solutions for the polynomial
equation
x^4 − 5x^3 − 30x^2 − 30x − 216 = 0.
1. x = −4, 9,−1
2. x = −4, 9
3. x = ±1, 6, 10
4. x = 1,−6, 10
5. x = −10, 6
6. None of these

anonymous · Answer

so let me make sure i understand the problem...

FInd all the real solutions for the polynomial equation
$$x^{4}-5x^{3}-30x^{2}-30x-216=0$$
1. x = -4, 9, -1
2. x = -4, 9
3. x = 1, -1, 6, 10
4. x = 1, -6, 10
5. x = -10, 6
6. None of these

is that correct?

anonymous · Answer

yes

anonymous · Answer

ok...

the best thing here would be to pick a value in an answer, try it and eliminate answers based on the result...

for instance, start with x = 1.

check to see if plugging in 1 for x in the equation gives you a zero

anonymous · Answer

$$(1)^{4}-5(1)^{3}-30(1)^{3}-30(1)-216$$
$$1-5-30-30-216$$
does that = 0?

anonymous · Answer

No.... 
this is what I got for the solutions 

x =  9 
x =  − 4 
x =  − √6·I 
x =  √6·I

anonymous · Answer

even better...so looking at just the real solutions there (there are two of them) which possible answer does that match?

anonymous · Answer

#2 but shouldn't it inclue - sqrt6*1?

anonymous · Answer

that's an i not a 1...those last 2 are imaginary pairs...

anonymous · Answer

yeah I meant to put i

anonymous · Answer

a fourth degree polynolmial has 4 solutions...

in this case, 2 of them are real solutions and the other 2 are imaginary...

all imaginary solutions come in conjugate pairs.

anonymous · Answer

$$i\sqrt{6}$$and $$-i\sqrt{6}$$ are not real solutions which is what was asked...so they wouldn't be in the list

anonymous · Answer

okay 
thanks