Find the coefficients of the polynomial...

Question

We have the polynomial $P(x)=x^4-3x^2-4$ and $Q(x)=x^2+mx+n$. Find the real coefficients of $m$ and $n$ , so that $P(x)$ is divisible to $Q(x)$.

Excuse for no signs of my work, the problem is I really do not how. Frankly, I've tried to divide the polynomials, but I think that's a stupid idea and the remainder should be zero, but this didn't work. So thank you in advance for your possible help!

In what sense did dividing not work? "The remainder should be zero" sounds like an equation you can solve. — , Feb 17 '14 at 16:43
well, both. Clearly, if polynomials divide each other, they share roots. — Karolis Juodelė, Feb 17 '14 at 16:43
Note that $Q(x)$ will divide $P(x)$ if the roots of $Q(x)$ are also roots of $P(x)$ (and multiple roots are multiple roots). $Q$ being quadratic has two roots. In the general case where the roots of $P$ are distinct (as here, in fact) there will be six different ways of picking an unordered pair of roots from $P$ to be the roots of $Q$. Whether these methods give you answers depends on the constraints on the coefficients. — Mark Bennet, Feb 17 '14 at 16:51

score 2 · Answer 1 · answered Feb 17 '14 at 16:43

2

Note that $$x^4 - 3x^2 - 4 = (\color{blue}{\bf x^2})^{\bf 2} - 3(\color{blue}{\bf x^2}) - 4 = (x^2 - 4)(x^2 + 1)= (x-2)(x+2)(x^2 + 1)$$

answered Feb 17 '14 at 16:43

amWhy

209,954

Also note $x^2-4=(x+2)(x-2)$ – Ali Caglayan Feb 17 '14 at 16:45
I appreciate your help, thank you!!! – wonderingdev Feb 17 '14 at 16:47
Yes, Alizter, indeed. – amWhy Feb 17 '14 at 16:48
You're welcome, John! – amWhy Feb 17 '14 at 16:48
@JohnG. Note that $(x^2+1)$ is irreducible over $\Bbb R$ so that means that the quadratic $Q$ must be comprised of the other factors. And just because I can in $\Bbb C$, $(x^2+1)=(x+i)(x-i)$ – Ali Caglayan Feb 17 '14 at 16:49

score 1 · Answer 2 · answered Feb 17 '14 at 16:43

1

HINT:

Using Middle Term factor,

$$z^2-3z-4=z^2-(4-1)z-4$$

Again, $$z^2-4z+z-4=(z-4)(z+1)$$

Set $z=x^2$

answered Feb 17 '14 at 16:43

lab bhattacharjee

274,582

YOur advise helped me a lot, thanks! – wonderingdev Feb 17 '14 at 16:46
@JohnG., nice to hear that – lab bhattacharjee Feb 17 '14 at 18:09

score 1 · Accepted Answer · answered Feb 17 '14 at 16:45

1

Hint: If $P(x)$ is divisible by $Q(x)$, then $Q(x)$ and $P(x)$ should have common roots.

answered Feb 17 '14 at 16:45

Hawk

6,540

Oooo, thank you very much now I understand – wonderingdev Feb 17 '14 at 16:46

score 0 · Answer 4 · edited Feb 17 '14 at 16:55

0

So how I understood we have $P(x)=(x^2-4)(x^2+1)$ So if $P(x)$ is divisible to $Q(x)$ so they share the same solutions, so:

$Q(-2)=4-2m+n=0$ and $Q(2)=4+2m+n=0$ So we find $m$ and $n$!!!

Is it right?

edited Feb 17 '14 at 16:55

Hawk

6,540

answered Feb 17 '14 at 16:51

wonderingdev

1,761

There are two solutions – Mark Bennet Feb 17 '14 at 16:55
Yes, 2 and -2 ! – wonderingdev Feb 17 '14 at 16:55
I think @MarkBennet meant two pairs of solutions – Hawk Feb 17 '14 at 16:57
May be I am wrong but x^2+1 has no solutions so we have one pair of solutions – wonderingdev Feb 17 '14 at 16:58
$x^2+1 \implies x=i,-i$ – Hawk Feb 17 '14 at 16:58
But the coefficients should be real?! – wonderingdev Feb 17 '14 at 16:59
I couldn't understand your last comment, however, what I understood you meant is that for $x=i,-i$ does not count for real $m,n$. But, $m,n$ are sum of products of roots respectively,which are real. – Hawk Feb 17 '14 at 17:02
I meant that the coefficients should not be complex numbers --from condition of the exercise – wonderingdev Feb 17 '14 at 17:04
Co-efficients are real when you consider the imaginary solution. As,$m=i+(-i)=0,n=i.(-i)=1$ – Hawk Feb 17 '14 at 17:05
I understand, thank you – wonderingdev Feb 17 '14 at 17:06
1

Easiest to see that $x^2+1$ is in the required form so must provide a solution. – Mark Bennet Feb 17 '14 at 17:07
@MarkBennet Very true. – Hawk Feb 17 '14 at 17:08

Find the coefficients of the polynomial...

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