Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need some help with the following problem. Many thanks in advance.

Let $f(x) = x^2+px+q$ and $g(x) = x^2+rx+s$. Find an expression for $f(g(x))$ and hence find a necessary and sufficient condition on $a$, $b$, $c$ for it to be possible to write the quartic expression $x^4+ax^3+bx^2+cx+d$ in the form $f(g(x))$, for some choice of values of $p$, $q$, $r$, and $s$.

Okay the first thing I did was to find $f(g(x))$:

$\begin{aligned}f(g(x)) & = (x^2+rx+s)^2+p(x^2+rx+s)+q \\& = x^2(x^2+rx+s)^2+rx(x^2+rx+s)+s(x^2+rx+s)+px^2+prx+ps+q \\& = x^4+rx^3+sx^2+rx^3+r^2+x^2+rsx+sx^2+srx+s^2+px^2+prx+ps+q \\& = x^4+(2r)x^3+(p+2s+r^2)x^2+(2rs+pr)x+s^2+q+ps \end{aligned}$

So we wish to have:

$x^4+(2r)x^3+({p}+2s+r^2)x^2+(2rs+pr)x+s^2+q+ps \equiv x^4+ax^3+bx^2+cx+d$

Comparing the coefficients, $r = \frac{1}{2}a$, $2s = b-\frac{1}{4}a^{2}-{p}$, and $c = r(2s+p)$, thus $c = \frac{ab}{2}-\frac{1}{{8}}a^{{3}} $.

I understand that this condition is 'necessary' -- my problem is that I'm not quite sure how to make it sufficient. I'm not quite sure how I'm supposed to choose some suitable values of p, q, r and s.

Show further that this condition holds if and only if it's possible to write the quartic expression $x^4+ax^3+bx^2+cx+d$ in the form $(x^2+vx+w)^2-k$, for some choice of values v, w, q, r, s.

$\begin{aligned} (x^2+vx+w)^2-k & = x^2(x^2+vx+w)+vx(x^2+vx+w)+w(x^2+vx+w)-k \\& = x^4+vx^3+wx^2+vx^3+vx^2+wvx+wx^2+vwx+w^2-k \\& = x^4+(2v)x^3+(2w+v^2)x^2+(2vw)x+w^2-k. \end{aligned}$

I see that the 'suitable choice' would have been q = 0, but how was I supposed to see that?

share|cite|improve this question
I don't think your ccondition on $c$ follows. $\frac{ab}{2} - \frac{1}{8}a^2$ is $2rs$, so you seem to be saying that $rp = -\frac{1}{8}a^2$, hence that $p=-\frac{1}{4}a$; on what grounds? Also, you never addressed $d$ or $q$. – Arturo Magidin Jun 17 '11 at 23:56
To prove your condition sufficient, you should be able to take any $a,b,c,d$ satisfying your condition(s) and derive $p,q,r,s$ – Ross Millikan Jun 18 '11 at 0:21
@Arturo Magidin. My apology! I've made a mistake when typing and omitted a p. I highlighted the change with blue fonts. As for considering d or q, the question asked me to find the condition on a, b, c. It then wants me to find a suitable value in p, q, r and s such that the condition is sufficient. Or at least that's how I understood it. – Lyrebird Jun 18 '11 at 0:21
Shouldn't the last term in your expression for $c$ be $-\frac{1}{8}a^3$? – Ross Millikan Jun 18 '11 at 0:27
So given $a,b,d$, you can calculate $p+2s,r,$ and $s^2+q+ps$. You have a degree of freedom left. As you say, you can pick $p$ and then calculate $q,s$. For the second part, you can just show that the coefficients in your expansion satisfy the same restriction on $a,b,c$. In the last line, it should be $k$, not $k^2$ – Ross Millikan Jun 18 '11 at 2:18

[A community wiki answer to remove this from the unanswered pool.]

To see that a condition is necessary and sufficient, you need to prove both directions. Since you want to find a condition on $\{a,b,c\}$ in order to do something involving choosing $\{p,q,r,s\}$, you should show that whenever the condition holds, you can do that something (i.e. there are always $\{p,q,r,s\}$ when the condition holds) and that whenever you can do that something, the condition must hold (i.e. given $\{p,q,r,s\}$ work you know the condition holds).

For the second part, you simply need to check that the conditions here are the same as the conditions above. (In fact, the same as either the condition on $\{a,b,c\}$ or the assumption that it can be expanded in $\{p,q,r,s\}$ - because you know these are equivalent.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.