Question

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 $f(1)?$

$-9009$

$-8008$

$-7007$

$-6006$

$-5005$

Solution:

We know that $f(x)$ has $4$ roots, $3$ of which are the roots of $g(x).$ This means that we can express $f(x)$ as $f(x) = g(x)(x - r),$ for some complex number $r$ that is the other root of $f(x).$

Plugging in $g(x),$ we get $f(x)$ equals: $(x^3 + ax^2 + x + 10)(x - r)$ $= x^4 + (a - r)x^3 + (1 - ar)x^2$ $+ (10 - r)x - 10r.$

Comparing coefficients, we get $10 - r = 100$ $r = -90.$ We also know that $a - r = 1$ $a = -89.$

Finally, we have that $f(1)$ equals: $1^4 + (a - r)1^3 + (1 - ar)1^2$ $+ (10 - r)1 - 10r$ $= 1+ (-89 + 90) + (1 - 89 \cdot 90)$ $+ (10 + 90) + 10 \cdot 90$ $= 1 + 1 - 8009 + 100 + 900$ $= -7007.$

Thus, C is the correct answer.

2017 AMC 10A Problem 24

Solution:

Problem 24 in Other Years