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Difference between revisions of "1965 AHSME Problems/Problem 19"
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== Solution 1 ==
 
== Solution 1 ==
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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 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
 +
 +
In f(x)
 +
 +
<math>r_1+r_2+r_3=-3</math>
 +
 +
<math>r_1r_2 + r_2r_3 + r_1r_3=9</math>
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 +
<math>r_1r_2r_3=-3</math>
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In g(x)
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 +
<math>r_1+r_2+r_3+r_4=-4</math>
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<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_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>p=2</math>
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<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+-1×9=-4q</math>
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<math>q=3</math>
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<math>r_1r_2r_3r_4=r</math>
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<math>-3×-1=r</math>
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<math>r=3</math>
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so <math>(p+q)r=\fbox{15}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 11:13, 26 March 2022

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 formula

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 $(p+q)r=\fbox{15}$

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
20
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All AHSME Problems and Solutions

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