If x^2 +4x +6 is a factor of x^4 +rx^2 +s, then r+s is:

One of us obtained 32, while the other got 52. I would just like to see what the correct answer is.

All exponents are all just the single number.

2. I never will understand this sort of dispute. This is not meteorology where certain signs may mean rain or they may not. One of you should be able to prove your answer. Who tried? Who succeeded?

Ignoring all the variables...
(1 0 r) - (1 4 6) = (0 -4 r-6)

(-4 r-6 0) - (-4 -16 -24) = (0 r+10 24)

(r+10 24 s) - (r+10 4r+40 6r+60) = (0 -4r-16 s-6r-60)

-4r-16 = 0
s-6r-60 = 0

Now what?

3. Originally Posted by jd3005
If x^2 +4x +6 is a factor of x^4 +rx^2 +s, then r+s is:

One of us obtained 32, while the other got 52. I would just like to see what the correct answer is.

All exponents are all just the single number.

Long division: $\displaystyle x^4+rx^2+s=(x^2+4x+6)(x^2-4x+(r+10))+\left[(-16-4r)x+(s-6r-60)\right]$ . The last part within the square parenthes4s is the residue and this must be zero, and from here that $\displaystyle -16-4r=0\,,\,\,s-6r-60=0$ , and solving this we get that $\displaystyle r+s=32$ .

Tonio

4. Let $\displaystyle x^4+rx^2+s = (ax^2+bx+c)(x^2+4x+6)$

Then $\displaystyle x^4+rx^2+s = ax^2(x^2+4x+6)+bx(x^2+4x+6)+c(x^2+4x+6)$

$\displaystyle \Rightarrow x^4+rx^2+s = ax^4+4ax^3+6ax^2+bx^3+4bx^2+6bx+cx^2+4cx+6c$

$\displaystyle \Rightarrow x^4+rx^2+s = ax^4+4ax^3+bx^3+6ax^2+4bx^2+cx^2+6bx+4cx+6c$

$\displaystyle \Rightarrow x^4+rx^2+s = (a)x^4+(4a+b)x^3+(6a+4b+c)x^2 +(6b+4c)x+6c$

Then compare the coefficients:

$\displaystyle a = 1$

$\displaystyle 4a+b = 0 \Rightarrow 4(1)+b = 0 \Rightarrow b = -4$

$\displaystyle 6a+4b+c = r \Rightarrow 6(1)+4(-4)+c = r \Rightarrow -10+c = r$

$\displaystyle 6b+4c = 0 \Rightarrow 6(-4)+4c = 0 \Rightarrow -24+4c = 0 \Rightarrow 4c = 24 \Rightarrow c = \frac{24}{4} = 6$

$\displaystyle s = 6c = 6(6) = 36$

$\displaystyle r = -10+c = -10+6 = -4.$

We have $\displaystyle s = 36$ and $\displaystyle r = -4$. Hence $\displaystyle s+r = 36-4 = 32$.