AMC 10 · 2019 · #14

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

complex-numberspolynomial-rootsvieta-formulasquadratic-equationssystematic-enumeration convert-to-algebrasystematic-enumerationcasework ↑ Prerequisites: complex-numberspolynomial-roots

📏 Medium solution 💡 3 insights

Problem

A degree-six polynomial is handed to us already split into three quadratic factors. The middle factor carries one unknown complex coefficient. The whole product has exactly four distinct roots. Find the modulus of that unknown coefficient.

Pick an answer.

(A)

(B)

$\sqrt{6}$

(C)

$2\sqrt{2}$

(D)

(E)

$\sqrt{10}$

How to solve

Strategy Organize Information in More Ways

As written, P is a list of coefficients, and coefficients say nothing about how many distinct roots there are. The same polynomial can be stored a second way: as the multiset of its six roots. That second storage format makes the given condition readable at a glance, because "exactly 4 distinct" is a statement about the root list, not about the coefficients. So the plan is to convert every factor we can into roots, see what the counting condition forces on the one factor we cannot, and then convert back to a coefficient using the sum-of-roots relation.

1STEP 1

Count roots with multiplicity first

With multiplicity there are six roots.

deg P = 2 + 2 + 2 = 6, #{roots with multiplicity} = 6, #{distinct roots} = 4

2STEP 2

Solve the two known quadratics

Solve the two known quadratics first.

x = (2 ± √(-4))/2 = 1 ± i, x = (4 ± √(-16))/2 = 2 ± 2i

3STEP 3

Stop chasing c, chase the roots

Chase the roots instead of the coefficient.

roots of x² - cx + 4 ⊆ {1+i, 1-i, 2+2i, 2-2i}

4STEP 4

Name the two roots and use Vieta

The root relations give the sum and product.

(x-r)(x-s) = x² - (r+s)x + rs ⟹ r + s = c, rs = 4

5STEP 5

Test every pair for product 4

Test every pair with the right product.

(1+i)(2-2i) = 2 - 2i + 2i + 2 = 4, (1-i)(2+2i) = 2 + 2i - 2i + 2 = 4

6STEP 6

Add the survivors, take the modulus

The modulus is root ten.

c = 3 - i or c = 3 + i, |c| = √(9 + 1) = √(10)

Answer

√(10)

Put c = 3 - i back in. Then x² - (3-i)x + 4 has root sum 3 - i and root product 4, which is exactly the pair 1+i and 2-2i. Both of those already appear among the roots of the other two factors, so the full root list of P is 1+i, 1-i, 2+2i, 2-2i, 1+i, 2-2i — six roots with multiplicity, four distinct values, as required. The check also confirms no case was missed: a middle factor with a repeated root outside the pool would give 5 distinct roots, and two new roots would give 6, so restricting to the pool lost nothing. Finally √(10) ≈ 3.16, and both candidate values of c have real part 3, so a modulus a little above 3 is exactly the size to expect.

💡Key takeaway

A polynomial is its list of roots in disguise, so when a question counts roots, switch to the root list and read the coefficients back off at the end.

Count roots with multiplicity first
Solve the two known quadratics
Stop chasing c, chase the roots
Name the two roots and use Vieta
Test every pair for product 4
Add the survivors, take the modulus