Difference between revisions of "1965 AHSME Problems/Problem 19"

Latest revision as of 10:17, 29 July 2024

Problem 19

If $x^4 + 4x^3 + 6px^2 + 4qx + r$ is exactly divisible by $x^3 + 3x^2 + 9x + 3$ , the value of $(p + q)r$ is:

$\textbf{(A)}\ - 18 \qquad \textbf{(B) }\ 12 \qquad \textbf{(C) }\ 15 \qquad \textbf{(D) }\ 27 \qquad \textbf{(E) }\ 45 \qquad$

Solution 1

Let $f(x)=x^3+3x^2+9x+3$ and $g(x)=x^4+4x^3+6px^2+4qx+r$ .

Let 3 roots of $f(x)$ be $r_1, r_2$ and $r_3$ . As $f(x)|g(x)$ , 3 roots of 4 roots of $g(x)$ will be same as roots of $f(x)$ . Let the 4th root of $g(x)$ be $r_4$ . By Vieta's Formulas

In $f(x)$

$r_1+r_2+r_3=-3$

$r_1r_2+r_2r_3+r_1r_3=9$

$r_1r_2r_3=-3$

In $g(x)$

$r_1+r_2+r_3+r_4=-4$

$=>r_4=-1$

$r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p$

$=>r_1r_2+r_1r_3+r_2r_3+r_4(r_1+r_2+r_3)=6p$

$=>9+(-1)(-3)=6p$

$=>p=2$

$r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q$

$=>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q$

$=>-3+(-1)(9)=-4q$

$=>q=3$

$r_1r_2r_3r_4=r$

$=>(-3)(-1)=r$

$=>r=3$

so $(p+q)r=\boxed{\textbf{(C) }15}$

By ~Ahmed_Ashhab

Solution 2

Notice that to obtain the $x^4$ term one must multiply $x^4+4x^3+6px^2+4qx+r$ by some linear function of the form $x-a$ . Looking at the $x^3$ term, it is clear that $a$ must equal $1$ . Therefore by multiplying $x^4+4x^3+6px^2+4qx+r$ by $x+1$ , the product will be $x^4+4x^3+12x^2+12x+3$ . Therefore $p=2$ , $q=3$ , $r=3$ . Thus $(2+3)3=\boxed{\textbf{(C) }15}$

1965 AHSC (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 == Solution 1 ==
+Let <math>f(x)=x^3+3x^2+9x+3</math> and <math>g(x)=x^4+4x^3+6px^2+4qx+r</math>.
+Let 3 roots of <math>f(x)</math> be <math>r_1, r_2 </math> and <math>r_3</math>. As  <math>f(x)|g(x)</math> , 3 roots of 4 roots of <math>g(x)</math> will be same as roots of <math>f(x)</math>. Let  the 4th root of <math>g(x)</math> be <math>r_4</math>.  By [[Vieta's Formulas]]
+In <math>f(x)</math>
+<math>r_1+r_2+r_3=-3</math>
+<math>r_1r_2+r_2r_3+r_1r_3=9</math>
+<math>r_1r_2r_3=-3</math>
+In <math>g(x)</math>
+<math>r_1+r_2+r_3+r_4=-4</math>
+<math>=>r_4=-1</math>
+<math>r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p</math>
+<math>=>r_1r_2+r_1r_3+r_2r_3+r_4(r_1+r_2+r_3)=6p</math>
+<math>=>9+(-1)(-3)=6p</math>
+<math>=>p=2</math>
+<math>r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q</math>
+<math>=>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q</math>
+<math>=>-3+(-1)(9)=-4q</math>
+<math>=>q=3</math>
+<math>r_1r_2r_3r_4=r</math>
+<math>=>(-3)(-1)=r</math>
+<math>=>r=3</math>
+so <math>(p+q)r=\boxed{\textbf{(C) }15}</math>
+By ~Ahmed_Ashhab
+== Solution 2 ==
+Notice that to obtain the <math>x^4</math> term one must multiply <math>x^4+4x^3+6px^2+4qx+r</math> by some linear function of the form <math>x-a</math>. Looking at the <math>x^3</math> term, it is clear that <math>a</math> must equal <math>1</math>. Therefore by multiplying <math>x^4+4x^3+6px^2+4qx+r</math> by <math>x+1</math>, the product will be <math>x^4+4x^3+12x^2+12x+3</math>. Therefore  <math>p=2</math>, <math>q=3</math>, <math>r=3</math>. Thus <math>(2+3)3=\boxed{\textbf{(C) }15}</math>
 == See Also ==
-{{AHSME 40p box|year=1965|num-b=39|after=Last Question}}
+{{AHSME 40p box|year=1965|num-b=18|num-a=20}}
 {{MAA Notice}}
 [[Category:Intermediate Algebra Problems]]

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Difference between revisions of "1965 AHSME Problems/Problem 19"

Latest revision as of 10:17, 29 July 2024

Contents

Problem 19

Solution 1

Solution 2

See Also