Another way to solve this? (1 Viewer)

totallybord · Mar 28, 2017

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

Thanks!

He-Mann · Mar 28, 2017

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 · Mar 28, 2017

$\therefore Q(x) = x^4 + px^3 +qx^2 -5x + 1 = (x^2 + 2)T(x) + (1-7x)\\ \\ \therefore Q(\sqrt 2 i) = 4 - 2\sqrt 2 pi -2q - 5\sqrt 2 i + 1 = 0 + (1 - 7i)\\ \\ \therefore (5-2q) - (2p+5)\sqrt 2 i = 1 - 7\sqrt 2 i$

Equating reals and imaginaries:

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

2p + 5 = 7 ==> p = 1

totallybord · Mar 28, 2017

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

Drongoski · Mar 28, 2017

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

Good point.

$Q(x) = (x^2 + 2)(x^2 + px + q-2) - (2p+5)x -2q + 5\\ \\ \therefore -(2p+5)x - 2q +5 = 1 - 7x\\ \\ \implies 2p+5 = 7 $ and $ -2q+5 = 1 \implies p=1 $ and $ q = 2$

totallybord · Mar 29, 2017

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

Drongoski · Mar 30, 2017

totallybord said:
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!

Mahan1 · Apr 1, 2017

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

View attachment 33856

Thanks!

$\textrm{well, long division is the fastest way but this is a way that does not use long division.}$

$Q(1) = 0 \Rightarrow p+q = 3$

$Q(x) = (x^{2} +2)T(x) +1-7x = 0 , \ \textrm{where} \ T(x) = ax+bx+c$

$Q(0) = 1 = 2T(0) +1 \Rightarrow T(0) = 0 \Rightarrow c= 0$

$Q(1) = 3T(1)-6 = 0 \Rightarrow T(1) = 2$

$\therefore a+b = 2$

$Q(-1) = 3T(-1) + 8 = 7 + q - p \Rightarrow 3T(-1) = q - p -1, \ \textrm{Note} \ T(-1) = a-b$

$3T(-1) + 3T(1) = 3(T(1) + T(-1) ) = 5 + q - p \Rightarrow 6a = 5 + q-p$

$\textrm{note, the coefficients of } \ x^{2} \ \textrm{in Q is} \ : 2a = q \Rightarrow 6a = 3q = 5 + q -p \Rightarrow 2q = 5-p \Rightarrow 2(q+p) = 6 = 5 + p$

$\therefore p =1.$

$\textrm{since} \ p+q =3 \Rightarrow q = 2.$

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Another way to solve this? (1 Viewer)

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