1975 AHSME Problems/Problem 13

Problem

The equation $x^6 - 3x^5 - 6x^3 - x + 8 = 0$ has

$\textbf{(A)} \text{ no real roots} \\ \textbf{(B)} \text{ exactly two distinct negative roots} \\ \textbf{(C)} \text{ exactly one negative root} \\ \textbf{(D)} \text{ no negative roots, but at least one positive root} \\ \textbf{(E)} \text{ none of these}$

Solution

Let $P(x) = x^6 - 3x^5 - 6x^3 - x + 8$ . When $x < 0$ , $P(x) > 0$ . Therefore, there are no negative roots.

Notice that $P(1) = -1$ and $P(0) = 8$ . There must be at least one positive root between 0 and 1, therefore the answer is $\boxed{\textbf{(D)}}$ .

1975 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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 • 26 • 27 • 28 • 29 • 30
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1975 AHSME Problems/Problem 13

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