2001 AMC 12 Problems/Problem 23

Problem

A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

$\text{(A) }\frac {1 + i \sqrt {11}}{2} \qquad \text{(B) }\frac {1 + i}{2} \qquad \text{(C) }\frac {1}{2} + i \qquad \text{(D) }1 + \frac {i}{2} \qquad \text{(E) }\frac {1 + i \sqrt {13}}{2}$

Solution

Let the polynomial be $P$ and let the two integer zeros be $z_1$ and $z_2$. We can then write $P(x)=(x-z_1)(x-z_2)(x^2+ax+b)$ for some integers $a$ and $b$.

If a complex number $p+qi$ with $q\not=0$ is a root of $P$, it must be the root of $x^2+ax+b$, and the other root of $x^2+ax+b$ must be $p-qi$.

We can then write $x^2+ax+b = (x-p-qi)(x-p+qi) = (x-p)^2 - (qi)^2 = x^2 - 2px + p^2 + q^2$.

We can now examine each of the five given complex numbers, and find the one for which the values $-2p$ and $p^2+q^2$ are integers. This is $\boxed{\frac {1 + i \sqrt {11}}{2}}$, for which we have $-2p = -2\cdot\frac 12 = -1$ and $p^2+q^2 = \left( \frac 12 \right)^2 + \left( \frac {\sqrt{11}}2 \right)^2 = \frac 14 + \frac {11}4 = \frac {12}4 = 3$.

(As an example, the polynomial $x^4 - 2x^3 + 4x^2 - 3x$ has zeroes $0$, $1$, and $\frac {1 \pm i \sqrt {11}}{2}$.)

Solution 2

By Vieta, we know that the product of all four zeros of the polynomial equals the constant at the end of the polynomial. We also know that the two imaginary roots are a conjugate pair (I.E if one is a+bi, the other is a-bi). So the two imaginary roots must multiply to give you an integer. Taking the 5 answers into hand, we find that $\boxed{\frac {1 + i \sqrt {11}}{2}}$ is our only integer giving solution.

Notes: this solution needs some justifications? or is it being wrong?

I think this solution is not correct. the products of the roots are integers do not mean the product of the two complex roots are integers.

Note: I believe this solution is correct. We know that the two real solutions are integers and that the final product is an integer. The product of the real solutions multiplied by the product of the complex solutions equals the final product, so we know that the product of the two complex solutions must be integers. We know the two complex solutions are conjugates, so we can test all the answer choices and find that A is the answer. ~A1597412

Solution 3

After dividing the polynomial out by $(x-p)$ and $(x-q)$, where p and q are the real roots of the polynomial, we will obtain a quadratic with two complex roots. We can then use the quadratic formula to solve for these complex roots.

Let's start by using synthetic division to divide $x^4+ax^3+bx^2+cx+d$ by $(x-p)$. Using this method, the quotient becomes $1x^3+(a-p)x^2+(b-pa+p^2)x+(c-bp+ap^2-p^3)+\frac{d-pc+bp^2-ap^3+p^4}{x-q}$. However, we know that there should be no remainder because $(x-p)$ is a factor of the polynomial, so $\frac{d-pc+bp^2-ap^3+p^4}{x-q}$ must equal 0, so $d=-pc+bp^2-ap^3+p^4$. When we divide the expression on the left by -p, we get $c-bp+ap^2-p^3$, so we can replace it in our original synthetic division equation with $\frac {d}{-k}$.

We then want to synthetically divide $x^3+(a-p)x^2+(b-pa+p^2)x+\frac {d}{-k}$ by the next factor, $(x-q)$. Using the same method as before, we can simplify the quotient to $x^2+(a-p-q)x+\frac{d}{pq}$. Now for the easy part!

Use the quadratic formula to determine the form of the complex roots.

$\frac{p+q-a\pm\sqrt{(a-p-q)^2-\frac{4d}{pq}}}{2}$

Now this is starting to look a lot like answers A and E. Noticing that the real part in each answer choice is $\frac{1}{2}$, $(k+n-a)=1$ and $(a-k-n)^2=1$, and the imaginary part is positive. Furthermore, by Vieta's Formulas, we know that d must be a multiple of p and q, so $\frac{4d}{pq}$ is a multiple of 4. Rearranging the expression, we get:

$\frac{1+i\sqrt{\frac{4d}{pq}-1}}{2}$

The radicand therefore must be one less than a multiple of four, which is only the case in $\frac {1 + i \sqrt {11}}{2}$ or $\boxed{A}$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS