Difference between revisions of "1980 AHSME Problems/Problem 24"

Latest revision as of 19:13, 16 August 2025

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

@@ Line 9: / Line 9: @@
 \text{(E)} \ 1.62  </math>
-== Solution ==
+== Solution 1 ==
-<math>\fbox{}</math>
+Since <math>(x-r)^2</math> is a factor of <math>8x^3-4x^2-42x+45</math>, <math>r</math> is a double root. Let <math>s</math> be the third root of <math>8x^3-4x^2-42x+45</math>.
+By Vieta's formulas, <math>2r+s = \frac{1}{2}</math>, so <math>s = \frac{1}{2} - 2r</math>.
+Also, by Vieta's formulas, <math>r^2 + 2rs = -\frac{21}{4}</math>.  Substituting <math>s = \frac{1}{2} - 2r</math> yields <math>r^2 + 2r\left(\frac{1}{2} - 2r\right) = -3r^2 + r = -\frac{21}{4}</math>.  So <math>12r^2 - 4r - 21 = 0</math>, that is, <math>(6r + 7)(2r - 3) = 0</math>.  Therefore, <math>(r, s)</math> is either <math>\left(\frac{3}{2}, -\frac{5}{2}\right)</math> or <math>\left(-\frac{7}{6}, \frac{17}{6}\right)</math>.
+Finally, by Vieta's formulas, <math>r^2s = -\frac{45}{8}</math>.  Though <math>(r, s) = \left(-\frac{7}{6}, \frac{17}{6}\right)</math> does not satisfy this equation, <math>\left(r, s\right) = \left(\frac{3}{2}, -\frac{5}{2}\right)</math> does.
+Therefore, <math>r = \frac{3}{2} = 1.5</math>, so <math>\boxed{\text{(\textbf{D})} \ 1.52}</math> is closest to <math>r</math>.
+-e_power_pi_times_i, edited by j314andrews
+== Solution 2 ==
+By polynomial long division, <math>\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}</math>.
+So <math>(24r^2 - 8r - 42)x + (-16r^3 + 4r^2 + 45) = 0</math>, that is, <math>r</math> is a root of both <math>24r^2 - 8r - 42</math> and <math>-16r^3 + 4r^2 + 45</math>.
+Since <math>24r^2 - 8r - 42 = 2(6r + 7)(2r - 3)</math>, either <math>r = -\frac{7}{6}</math> or <math>r = \frac{3}{2}</math>.  By the Rational Root Theorem, <math>-\frac{7}{6}</math> is not a root of <math>-16r^3 + 4r^2 + 45</math>.  However, <math>-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</math>, so <math>\frac{3}{2}</math> is a root of <math>-16r^3 + 4r^2 + 45</math>.
+Therefore, <math>r = \frac{3}{2} = 1.5</math>, so <math>\boxed{(\textbf{D})\ 1.52}</math> is closest to <math>r</math>.
+-j314andrews
+== Solution 3 ==
+Let <math>y = 2x</math>.  Then <math>x = \frac{y}{2}</math>, so <math>8x^3 - 4x^2 - 42x + 45 = y^3 - y^2 - 21y + 45</math>.
+By the Rational Root Theorem, any rational root of <math>y^3 - y^2 - 21y + 45</math> is a factor of <math>45</math>.  Of the factors of <math>45</math>, <math>3</math> and <math>5</math> are roots, and <math>y^3 - y^2 - 21y + 45 = (y-3)^2(y-5)</math>.  So <math>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)</math>, and <math>r = \frac{3}{2} = 1.5</math>. So <math>\boxed{(\textbf{D})\ 1.52}</math> is closest to <math>r</math>.
+-j314andrews
 == See also ==

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