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 October 14th, 2011, 09:32 PM #1 Member   Joined: Oct 2011 Posts: 81 Thanks: 0 Value of two variables If $x^2 + 2x +5$ is a factor of $x^4 + px^2 + q$. Find the values of p and q. How to do this question?
 October 14th, 2011, 09:58 PM #2 Senior Member   Joined: Aug 2008 Posts: 113 Thanks: 0 Re: Value of two variables Let $x^4 + px^2 + q= (x^2 +2x + 5)*Q(x)$ Note that $x= -1 \pm 2i$ are the two roots of $x^2 +2x + 5= 0$ Put $x= -1 + 2i$ in the above equation. $(-1 + 2i)^2= -3 - 4i$ $(-1 + 2i)^4= (-3 - 4i)^2 = -7 + 24i$ So $-7 + 24i + p(-3 - 4i) + q= 0$ Or, $-3p + q - 7= 0$ and $-4p + 24= 0$ So $p= 6$ and $q= 25$ ----------------------------------------------- Verification: Put $x= -1 - 2i$ in the above equation: $(-1 - 2i)^2= -3 + 4i$ $(-1 - 2i)^4= (-3 + 4i)^2 = -7 - 24i$ $-7 - 24i + p(-3 + 4i) + q= 0$ $-3p + q - 7= 0 and 4p - 24 = 0$ So we get the same values of p and q.
 October 15th, 2011, 02:07 AM #3 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Re: Value of two variables $x^4+px^2+q=\left(x^2+2x+5\right).(x^2+ax+b)$ Where $5b=q$
 October 16th, 2011, 10:32 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,034 Thanks: 2271 Code:               x² - 2x  +     p-1 ------------------------------------ x² + 2x + 5 x^4 + px²           +    q              x^4 + 2x³ + 5x²              -------------------                  - 2x³ + (p-5)x²                  - 2x³ - 4x² -     10x                  -------------------------                          (p-1)x² +     10x + q                          (p-1)x² + 2(p-1)x + 5p-5                          ------------------------ Hence p = 6 and q = 25. Alternatively, notice that the second factor must be x² - 2x + 5 to make the product an even function, then expanding (x² + 2x + 5)(x² - 2x + 5) gives x^4 + 6x² + 25, so p = 6 and q = 25.

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