1954 AHSME Problems/Problem 20

Problem 20

The equation $x^3+6x^2+11x+6=0$ has:

$\textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root}$

Solution

By the rational root theorem, $1, -1, 2, -2, 3, -3, 6, -6$ are possible rational roots. Because $x^3+6x^2+11x+6>0$ for $x>0$ , so there are no positive roots. We try $-1, -2, -3, -6$ , so $x=-1, x=-3, x=-2$ , so there are no positive real roots; $\fbox{B}$

Solution 2

By Descartes' Rule of Signs, there are no sign changes (all coefficients of terms are positive), so there are no positive real roots, or $\fbox{B}$ .

1954 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1954 AHSME Problems/Problem 20

Contents

Problem 20

Solution

Solution 2

See Also