Quick question (1 Viewer)

[Aesytic]

a polynomial P(x) has equation P(x) = ax^3 + bx^2 + cx + d where a, b, c and d are real

I. the polynomial must cut the x-axis at least once
II. if P(alpha) = 0, then P(conj. alpha)=0

are both these statements correct?

 

[lolcakes52]

The first is definitely correct! From our examination of polynomials we know that the limit as x approaches positive or negative infinity is ax^3. So we have a change is sign from positive to negative and hence at least one real root! The second is also correct but its harder to explain why. But basically the polynomial can be written as P(x)=(x-(a+ib))Q(x) which only has real coefficients if (x-(a-ib)) is also a factor.

 

[Aesytic]

yeah, i thought both were correct too. but what if a,b and c were all 0? wouldn't that mean that the first statement wouldn't be correct?

 

[lolcakes52]

Do you mean a,b,c and d? If you didn't then no as the Polynomial is just P(x)=d which is has a y value of d no matter what the value of x is. However, for the case where all the coefficients are equal to zero, P(x) is the zero polynomial which technically cuts the x axis infinitely and for all values of x, complex or not, so both statements are true for the this case.

 

[Aesytic]

ok, so the first statement is correct unless the polynomial is P(x)=d where d isn't 0, and the second statement is correct no matter what?

 

[lolcakes52]

Both statements are correct if a,b,c and d are zero. Neither if a,b and c are but is not.

 

[Aesytic]

ah, alright. thanks for that!

 

[jyu]

'I. the polynomial must cut the x-axis at least once' is correct if we consider x as real. So 'II. if P(alpha) = 0, then P(conj. alpha)=0' is not correct.
II is correct if we consider x as complex. So I is not correct.
.: the statements are mutually exclusive.

 

[barbernator]

'I. the polynomial must cut the x-axis at least once' is correct if we consider x as real. So 'II. if P(alpha) = 0, then P(conj. alpha)=0' is not correct.
II is correct if we consider x as complex. So I is not correct.
.: the statements are mutually exclusive.

wut? the statements are not mutually exclusive. by definition, a real number is also a complex number, so we are always considering x as complex

 

[jyu]

I am referring to the statements, not the numbers.

 

[barbernator]

I am referring to the statements, not the numbers.

i don't totally get you, but i guess if he rephrased as

I) the polynomial must have at least 1 real zero
II) the unreal factors occur in conjugate pairs

it would better suit

 

[Aesytic]

it was a question in my half-yearly
that was the information given, and it was a multiple choice question asking which of the statements were correct
A. I only
B. II only
C. Both
D. Neither

 

[jet]

II is true when the coefficients are real. Why? Let be a root.



If the coefficients weren't real, then and the proof would fail.

 

[Aesytic]

yeah, i understand why II is correct, but i wasn't exactly sure about I because although P(x) suggests that it's a cubic, if a was 0, then it wouldn't be so anymore, and if a, b and c were 0 then the polynomial wouldn't actually cut the x-axis provided d isn't 0