Difference between revisions of "1960 AHSME Problems/Problem 31"

(Solution to Problem 31)
 
m (Solution)
Line 21: Line 21:
 
Substitute <math>c</math> and <math>b</math> back to get
 
Substitute <math>c</math> and <math>b</math> back to get
 
<cmath>x^4 + 6x^2 + 25</cmath>
 
<cmath>x^4 + 6x^2 + 25</cmath>
Thus, <math>p=6</math> and <math>q=25</math>, so the answer is <math>\boxed{\textbf{(D)}}</math>
+
Thus, <math>p=6</math> and <math>q=25</math>, so the answer is <math>\boxed{\textbf{(D)}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=30|num-a=32}}
 
{{AHSME 40p box|year=1960|num-b=30|num-a=32}}

Revision as of 18:44, 12 May 2018

Problem

For $x^2+2x+5$ to be a factor of $x^4+px^2+q$, the values of $p$ and $q$ must be, respectively:

$\textbf{(A)}\ -2, 5\qquad \textbf{(B)}\ 5, 25\qquad \textbf{(C)}\ 10, 20\qquad \textbf{(D)}\ 6, 25\qquad \textbf{(E)}\ 14, 25$

Solution

Let the other quadratic be $x^2 + bx + c$, where $(x^2 + 2x + 5)(x^2 + bx + c) = x^4 + px^2 + q$. Multiply the two quadratics to get \[x^4 + (b+2)x^3 + (c + 2b + 5)x^2 + (2c + 5b)x + 5c\] Since $x^4 + px^2 + q$ have no $x^3$ term and no $x$ term, the coefficients of these terms must be zero. \[b+2=0\] \[b=-2\] \[2c+5b=0\] \[2c-10=0\] \[c=5\] Substitute $c$ and $b$ back to get \[x^4 + 6x^2 + 25\] Thus, $p=6$ and $q=25$, so the answer is $\boxed{\textbf{(D)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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