2010 AMC 12A Problems/Problem 21

Problem 21

The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

Solution

The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$ .

Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$ .

Suppose we let $p$ , $q$ , and $r$ be the roots of this function, and let $x^3-ux^2+vx-w$ be the cubic polynomial with roots $p$ , $q$ , and $r$ .

$(x-p)(x-q)(x-r) = x^3-ux^2+vx-w$

$(x-p)^2(x-q)^2(x-r)^2 = x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0$

$\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-ux^2+vx-w = 0$

In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}$ we must first expand out the terms of $(x^3-ux^2+vx-w)^2$ .

$(x^3-ux^2+vx-w)^2$ $= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2$

[Quick note: Since we don't know $a$ , $b$ , and $c$ , we really don't even need the last 3 terms of the expansion.]

$2u = 10$

$u^2+2v = 29$

$2uv+2w = 4$

$u = 5$

$v = 2$

$w = -8$

$\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8$

All that's left is to find the largest root of $x^3-5x^2+2x+8$ .

$x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)$

$\boxed{\textbf{(A)}\ 4}$

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2010 AMC 12A Problems/Problem 21

Problem 21

Solution