possible for cubic fcn to have only 2 real roots of order 1?

[ahlingg]

Is it possible for a cubic function to have only 2 real distinct roots, each of order 1? Explain why or why not. Include examples.

Please, and thank you :lol:

 

[pka]

Re: Polynomial Function

Complex roots of polynomials over the real field occur in pairs.
If \(\displaystyle z\) is a root of \(\displaystyle P(x)\) then so is \(\displaystyle \overline{z}\), z conjunct.

But what are the roots of \(\displaystyle x(x-2)^2\)?

 

[ahlingg]

Re: Polynomial Function

Hm, I'm sorry, what ? I'm not fully understanding what you're trying to say. (but that's my fault, of course)

The roots are
x=0, and
x=2 (equal roots) < but I guess the question is asking when there aren't equal roots ..

I'm more on the "No" side, but I can't seem to explain myself in an um , I guess you could say, more "intelligent" way .

 

[pka]

Re: Polynomial Function

The polynomial of degree three, \(\displaystyle x(x-2)^2\), has exactly two real roots \(\displaystyle \{0,2\}\).
So rethink you inclination.

 

[ahlingg]

Re: Polynomial Function

But 2 is an order 2 , and so therefore, it won't satisfy what the question is asking in the first place.

2 real distinct roots, each of order 1

 

[pka]

Re: Polynomial Function

2 real distinct roots, each of order 1

So the answer is NO.

 

[daon]

Re: Polynomial Function

As already said, complex roots come in conjugate pairs (To prove, let a+ib be a root with b not equal to zero and show that a-ib is also one). Moreover, by the fundamental theorem of algebra, all complex polynomials have their roots in C. If there were only two real, distinct roots of multiplicity 1 then you've contradicted one of these facts.