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

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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 <math>f(x)=x^3+3x^2+9x+3</math> and <math>g(x)=x^4+4x^3+6px^2+4qx+r</math>.  
  
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Let 3 roots of f(x) 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 g(x) will be same as roots of f(x). Let  the 4th root of g(x) be <math>r_4</math>.  By vieta's formula
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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]]
  
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In f(x)
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In <math>f(x)</math>
  
 
<math>r_1+r_2+r_3=-3</math>
 
<math>r_1+r_2+r_3=-3</math>
  
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<math>r_1r_2 + r_2r_3 + r_1r_3=9</math>
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<math>r_1r_2+r_2r_3+r_1r_3=9</math>
  
 
<math>r_1r_2r_3=-3</math>
 
<math>r_1r_2r_3=-3</math>
  
−
In g(x)
+
In <math>g(x)</math>
  
 
<math>r_1+r_2+r_3+r_4=-4</math>
 
<math>r_1+r_2+r_3+r_4=-4</math>
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<math>=>r_4=-1</math>
 
<math>=>r_4=-1</math>
  
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<math>r_1r_2 + r_1r_3 + r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p</math>
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<math>r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p</math>
  
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<math>r_1r_2 + r_1r_3 + r_2r_3+r_4(r_1+r_2+r_3)=6p</math>
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<math>=>r_1r_2+r_1r_3+r_2r_3+r_4(r_1+r_2+r_3)=6p</math>
  
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<math>9+-1×-3=6p</math>
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<math>=>9+(-1)(-3)=6p</math>
  
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<math>p=2</math>
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<math>=>p=2</math>
  
 
<math>r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q</math>
 
<math>r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q</math>
  
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<math>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q</math>
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<math>=>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q</math>
  
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<math>-3+-1×9=-4q</math>
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<math>=>-3+(-1)(9)=-4q</math>
  
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<math>q=3</math>
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<math>=>q=3</math>
  
 
<math>r_1r_2r_3r_4=r</math>
 
<math>r_1r_2r_3r_4=r</math>
  
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<math>-3×-1=r</math>
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<math>=>(-3)(-1)=r</math>
  
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<math>r=3</math>
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<math>=>r=3</math>
  
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so <math>(p+q)r=\fbox{15}</math>
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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 ==
 
== See Also ==
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{{AHSME 40p box|year=1965|num-b=19|after=20}}
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{{AHSME 40p box|year=1965|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

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 $(2+3)3=\boxed{\textbf{(C) }15}$

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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