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1980 AHSME Problems/Problem 24

Problem

For some real number $r$, the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$. Which of the following numbers is closest to $r$?

$\text{(A)} \ 1.22 \qquad \text{(B)} \ 1.32 \qquad \text{(C)} \ 1.42 \qquad \text{(D)} \ 1.52 \qquad \text{(E)} \ 1.62$

Solution 1

Since $(x-r)^2$ is a factor of $8x^3-4x^2-42x+45$, $r$ is a double root. Let $s$ be the third root of $8x^3-4x^2-42x+45$.

By Vieta's formulas, $2r+s = \frac{1}{2}$, so $s = \frac{1}{2} - 2r$.

Also, by Vieta's formulas, $r^2 + 2rs = -\frac{21}{4}$. Substituting $s = \frac{1}{2} - 2r$ yields $r^2 + 2r\left(\frac{1}{2} - 2r\right) = -3r^2 + r = -\frac{21}{4}$. So $12r^2 - 4r - 21 = 0$, that is, $(6r + 7)(2r - 3) = 0$. Therefore, $(r, s)$ is either $\left(\frac{3}{2}, -\frac{5}{2}\right)$ or $\left(-\frac{7}{6}, \frac{17}{6}\right)$.

Finally, by Vieta's formulas, $r^2s = -\frac{45}{8}$. Though $(r, s) = \left(-\frac{7}{6}, \frac{17}{6}\right)$ does not satisfy this equation, $\left(r, s\right) = \left(\frac{3}{2}, -\frac{5}{2}\right)$ does.

Therefore, $r = \frac{3}{2} = 1.5$, so $\boxed{\text{(\textbf{D})} \ 1.52}$ is closest to $r$.

-e_power_pi_times_i, edited by j314andrews

Solution 2

By polynomial long division, $\frac{8x^3 - 4x^2 - 42x + 45}{x^2 - 2rx + r^2} = 8x + (16r-4) + \frac{(24r^2 - 8r - 42)x + (-16r^3 + 4r^2 + 45)}{x^2 - 2rx + r^2}$.

So $(24r^2 - 8r - 42)x + (-16r^3 + 4r^2 + 45) = 0$, that is, $r$ is a root of both $24r^2 - 8r - 42$ and $-16r^3 + 4r^2 + 45$.

Since $24r^2 - 8r - 42 = 2(6r + 7)(2r - 3)$, either $r = -\frac{7}{6}$ or $r = \frac{3}{2}$. By the Rational Root Theorem, $-\frac{7}{6}$ is not a root of $-16r^3 + 4r^2 + 45$. However, $-16\left(\frac{3}{2}\right)^3 + 4\left(\frac{3}{2}\right)^2 + 45 = -16 \cdot \frac{27}{8} + 4 \cdot \frac{9}{4} + 45 = -54 + 9 + 45 = 0$, so $\frac{3}{2}$ is a root of $-16r^3 + 4r^2 + 45$.

Therefore, $r = \frac{3}{2} = 1.5$, so $\boxed{(\textbf{D})\ 1.52}$ is closest to $r$.

-j314andrews

Solution 3

Let $y = 2x$. Then $x = \frac{y}{2}$, so $8x^3 - 4x^2 - 42x + 45 = y^3 - y^2 - 21y + 45$.

By the Rational Root Theorem, any rational root of $y^3 - y^2 - 21y + 45$ is a factor of $45$. Of the factors of $45$, $3$ and $5$ are roots, and $y^3 - y^2 - 21y + 45 = (y-3)^2(y-5)$. So $8x^3 - 4x^2 - 42x + 45 = (2x-3)^2(2x-5) = 8\left(x-\frac{3}{2}\right)^2\left(x-\frac{5}{2}\right)$, and $r = \frac{3}{2} = 1.5$. So $\boxed{(\textbf{D})\ 1.52}$ is closest to $r$.

-j314andrews

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AHSME Problems and Solutions

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