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.

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.

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

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 $\boxed{A}$ .

2001 AMC 12 (Problems • Answer Key • Resources)
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
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2001 AMC 12 Problems/Problem 23

Contents

Problem

Solution

Solution 2

Solution 3

See Also