Another way to solve this? (1 Viewer)

totallybord

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So I've always done the following question using long division and then equated it. Is there another method?

Polynomials.PNG

Thanks!
 

He-Mann

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Goal: find p and q.

Information:
-Q(x) has a zero at x = 1; what does this imply?
-Q(x) has a remainder of (1-7x) when divided by (x^2 + 2); what does this imply?
 

Drongoski

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Equating reals and imaginaries:

5-2q = 1 ==> q = 2

2p + 5 = 7 ==> p = 1
 
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totallybord

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This was a 3U questions so no imaginary numbers are included- thanks though!
 

totallybord

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Hi Drongoski
How did you get the first line (without long division)?
 

Drongoski

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Hi Drongoski
How did you get the first line (without long division)?
What I have done is equivalent to long division. By inspection, I fiddled with the terms of the quartic polynomial like this: the 1st 3 terms x^4 + px^3 + qx^2 = x^2 (x^2 + px + q) = (x^2 + 2)(x^2 + px +q) - 2x^2 - 2px - 2q

Therefore Q(x) = the last expression - 5x + 1 = (x^2+2)(x^2+px+q) - 2x^2 - (2p+5)x - 2q + 1 = (x^2 +2)(x^2+px+q) -(x^2+2)(2) - (2p+5)x - 2q + 5
= (x^2+2)(x^2+px+q-2) - (2p+5)x - 2q + 5

Rather messy. You have to be very manipulative!
 
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