2017 AMC 10A Problem 24

P21-25~6 minPrint

Polynomials Set up the equation/formula and compute; no special trick needed Test small/specific values or special cases to find or verify the answer

For certain real numbers $a$ , $b$ , and $c$ , the polynomial $g(x) = x^3 + ax^2 + x + 10$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $f(x) = x^4 + x^3 + bx^2 + 100x + c.$ What is $f(1)$ ?

Show solution without answering

Key insight

Since f has all three roots of g, f equals g times (x - r). Matching x terms gives r = -90, so f(1) = g(1) times 91.

Solution

1. Why $f$ is $g$ times something. Call the roots of $g$ by $r_1, r_2, r_3$ . Since $g$ starts with $x^3$ , it factors as $g(x) = (x - r_1)(x - r_2)(x - r_3)$ . These three roots are also roots of $f$ , which starts with $x^4$ and so has one more root, call it $r$ . So
$f(x) = (x - r_1)(x - r_2)(x - r_3)(x - r) = g(x)\,(x - r).$
2. Multiply it out, keeping track of each power of $x$ :
$(x^3 + ax^2 + x + 10)(x - r) = x^4 + (a - r)x^3 + (1 - ar)x^2 + (10 - r)x - 10r.$
3. Match with $f(x) = x^4 + x^3 + bx^2 + 100x + c$ , one power at a time. Start with the easiest line:
- $x$ terms: $10 - r = 100$ , so $r = -90$ .
- $x^3$ terms: $a - r = 1$ , so $a = 1 + r = -89$ .
4. Find $f(1)$ without $b$ or $c$ . Put $x = 1$ into the factored form $f(x) = g(x)(x - r)$ :
$g(1) = 1 + a + 1 + 10 = 12 + (-89) = -77,$
$1 - r = 1 + 90 = 91,$
$f(1) = (-77)(91) = -7007.$

The answer is $\boxed{\textbf{(C)}\ -7007}$ .

Why this works

If every root of one polynomial is also a root of a bigger one, the bigger one is the smaller one times an extra factor. Writing that out and matching the numbers in front of each power of $x$ turns the problem into a few small equations. Pick the easiest equation first; here the $x$ term gave $r$ right away.

Alternative approach

Finish the matching and plug in. The $x^2$ terms give $b = 1 - ar = 1 - (-89)(-90) = 1 - 8010 = -8009$ . The constant gives $c = -10r = 900$ . Then
$f(1) = 1 + 1 + (-8009) + 100 + 900 = -7007.$

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since the three distinct roots of $g$ are roots of $f$ , the cubic $g(x)$ divides $f(x)$ . Both are monic, so the quotient is monic and linear:
$f(x) = g(x)(x - r) = (x^3 + ax^2 + x + 10)(x - r)$
for some real $r$ .

Expand and match coefficients with $x^4 + x^3 + bx^2 + 100x + c$ :
- $x^3$ : $a - r = 1$ .
- $x$ : $10 - r = 100$ , so $r = -90$ .
- $x^2$ : $1 - ar = b$ ; constant: $-10r = c$ .

From $r = -90$ we get $a = 1 + r = -89$ . We do not need $b$ or $c$ , because
$f(1) = g(1)\,(1 - r) = (1 + a + 1 + 10)(1 + 90) = (12 - 89)(91) = (-77)(91) = -7007.$

The answer is $\boxed{\textbf{(C)}\ -7007}$ .

Review in the AoPS books

Intermediate Algebra, Chapter 6: Polynomial Division (§6.2, §6.4)
Intermediate Algebra, Chapter 7: Polynomial Roots Part I (§7.1, §7.2, §7.3)
Intermediate Algebra, Chapter 8: Polynomial Roots Part II (§8.3)