# Paranoid Polynomial Problem (1 Viewer)

#### mecramarathon

##### Member
How the heck do you solve this?

A polynomial Q(x) = x^4 + px^3 + qx^2 - 5x + 1 has a zero at x = 1. When Q(X) is divided by x^2 + 2 it has a remainder of 1 - 7x. Find p and q.

if u solve it

#### daryl-d

##### Member
first use the remainder theorem, i.e. Q(1)=o, or (x-1) is a factor of Q(x)

therefore: 1+p+q -5 +1 =0

p+q=3

Then using longdivision:

u get Q(x) = (x^2 + 2) B(x) +R [ where R is the remainder and B(x) = x^2 +px +(q-2)] ]

R= -x(5+2p) -2q +1 [now equate remainders]

R=1-7x {given}

therefore: 1+2p=7 and -2q +5=1

hence p= 1 and q=2 {note this also satisfies p+q=3}

hope this helps i am a bit rusty with this

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#### daryl-d

##### Member
can someone please tell me if this is correct:

Thanks for the post below guys, i had a feeling my long divison was wrong

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#### cutemouse

##### Account Closed
(p, q) = (1, 2).

BTW, where's this question from?

EDIT: I've attached my solution. Where's it from though? I recall seeing it somewhere...

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#### Drongoski

##### Well-Known Member
Surprisingly I didn't even need to use fact x=1 is a zero. But I need to use complex numbers

$\bg_white \100dpi Q(x)\ =\ x^{4}+px^{3}+qx^{2}-5x+1\\ \\Since\ upon\ division\ by\ x^{2}+2 \ remainder\ is\ 1-7x:\\ \\Q(x)\ =\ (x^{2}+2)h(x)\ +\(1-7x)\\ \\ \therefore \ Q(\sqrt{2}i)\ =\ (\sqrt{2}i)^{4}+p(\sqrt{2}i)^{3}+q(\sqrt{2}i)^{2}-5(\sqrt{2}i)+1\ =\ 1-7\sqrt{2}i\\ \\ \therefore \ (5-2q)\ -\ (2\sqrt{2}p+5\sqrt{2})i\ =\ 1-7\sqrt{2}i\\ \\ Equating\ reals\ and \ imaginary:\\ \\ 5-2q\ =\ 1\ \ \ \Rightarrow \ \ q\ =\ 2\\ \\ 2\sqrt{2}p+5\sqrt{2}\ =\ 7\sqrt{2}\ \ \ \Rightarrow \ \ \ p\ =\ 1$

This approach not suitable for 3U-only students.

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#### cutemouse

##### Account Closed
Yeah, my way would be the 3U approach.