# Australian HSC mathematics extension 2 exam Polynomial

1. Oct 27, 2009

### Mentallic

1. The problem statement, all variables and given/known data
This problem is from the Australian HSC mathematics extension 2 exam. Q6b)

It states:

b) Let $$P(x)=x^3+qx^2+qx+1$$, where q is real. One zero of P(x) is -1

i) Show that if $\alpha$ is a zero of P(x) then $$\frac{1}{\alpha}$$ is a zero of P(x)

ii) Suppose that $\alpha$ is a zero of P(x) and $\alpha$ is not real.
(1) show that $$|\alpha|=1$$
(2) show that $$Re(\alpha)=\frac{1-q}{2}$$

3. The attempt at a solution

i) I was able to answer this one

ii) (1) I was uncertain about this one so I fudged the answer, but for a quick and unsatisfactory answer, I went with the idea that if $\alpha$ and $1/\alpha$ are complex roots to a real polynomial, then the roots must occur in conjugate pairs and thus have the same modulus. Actually... also

$$P(x)=x^3+qx^2+qx+1\equiv (x+1)(x^2+(q-1)x+1)$$

The quadratic factor: $$x=\frac{1-q\pm\sqrt{(q-3)(q+1)}}{2}$$
Thought I had it... guess not...

ii) (2) I scabbed the answer from the quadratic (without really understanding why it works) that $$Re(z)=Re(\alpha)=\frac{1-q}{2}$$
This is the part without the surd. Also, unsure of a correct approach to the answer.

Help for these 2 parts would be greatful.

2. Oct 27, 2009

### Office_Shredder

Staff Emeritus
Re: Polynomial

For part ii, the idea that they have to come in conjugate pairs sounds good to me. What did you find wrong with it?

For the last part notice $$x=\frac{1-q\pm\sqrt{(q-3)(q+1)}}{2}$$ is real unless the part inside the square root is negative. If that's negative, what's the real part of the number?

3. Oct 27, 2009

### Mentallic

Re: Polynomial

for ii) I thought that while I was able to show that$$|\alpha|=|\frac{1}{\alpha}|$$, I wasn't able to show they're both equal to 1. But now that I look at it again, for a complex number $\alpha$ and its reciprocal to be equal, they must have a mod of 1.

for iii) well yes the real part is ofcourse obvious, but this assumes that $$(q-3)(q+1)<0$$ thus $$-1<q<3$$
How do I explain/show that q is in fact in this range? Or can I just assume it because the question specifically stated that the roots are imaginary?

4. Oct 27, 2009

### Office_Shredder

Staff Emeritus
Re: Polynomial

If it's not in that range, then you've found all the roots, and they are real. But one of them has to be imaginary, so that's a contradiction

5. Oct 27, 2009

### Mentallic

Re: Polynomial

I understand now. Thanks for your help Office_Shredder