Paranoid Polynomial Problem (1 Viewer)

[mecramarathon]

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]

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

 

[daryl-d]

can someone please tell me if this is correct:


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

 

[cutemouse]

(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...

 

[Drongoski]

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






This approach not suitable for 3U-only students.

 

[cutemouse]

Yeah, my way would be the 3U approach.