# How do I go about algebraic manipulation of polynomials with many terms?

I'm doing an inductive proof for a homework problem, and for one step, I need to show that $$\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30} + (n+1)^4 = \\ \dfrac{(n+1)((n+1)+1)(2(n+1)+1)(3(n+1)^2+3(n+1)-1)}{30}$$

Of course, this can be easily verified with Wolfram Alpha. But to make the proof clear, I'd like to show the steps for arriving at the right side from the left side.

I've had to do this sort of thing before, and I eventually gave up and just started from the right side and worked my way backwards. This, of course, yielded steps that were totally non-obvious and thus unhelpful in the forward direction.

So, what's the legitimate way to do algebraic manipulation like this? I'm not asking for a worked out solution of this particular problem, but are there any general techniques that I'm not aware of? I pretty much have no idea where to start. I've considered factoring by grouping and all that jazz, but I'm just not seeing a good entry point here.

Any help would be much appreciated!

• If you have an identity of this kind that you’d like to verify, simply expand each side to a polynomial in $n$, with no parentheses whatever. Then check that the two polynomials that you’ve gotten are equal. – Lubin Jan 23 '14 at 2:17
• @Lubin, note that you can omit common factors that appear on both sides in order to reduce the work. Here, you can leave the $(n+1)$ as a factor instead of distributing it over the rest since the same factor appears on the other side already. You could also use linear factors on one side to try to guess at factorizations of the other side, thus working a little bit backwards from purely simplifying. – MPW Jan 23 '14 at 4:39
• the leading term is $6n^5$ on the lhs but $4n^5$ on the rhs. The constant term is $1$ on the lhs and $-1$ on the rhs. Your equation still does not hold. – miracle173 Jan 23 '14 at 5:38
• Gah, I am really bad at typing out long equations. I fixed it once more, and triple checked it this time. You're gonna like the new equation; I guarantee it. – Trent Bing Jan 25 '14 at 8:24

I am sure if I understand you right but I try to answer. I assume you want to prove the formula about the sum of the 4th power of the first n numbers and to show $$\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} + (n+1)^4 = \tag{1}\\ \frac{(n+1)((n+1)+1)(2(n+1)+1)(3(n+1)^2+3(n+1)-1)}{30}$$

In a comment @Lubin already suggested the way to handle this.

Start with the lhs (left hand side) of $(1)$ and expand:

$$\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} + (n+1)^4 =\\ \frac{n(2n^2+3n+1)(3n^2+3n-1)}{30} +(n^4+4n^3+6n^2+4n+1)=\\ \frac{n(6n^4+15n^3+10n^2 - 1)}{30} + (n^4+4n^3+6n^2+4n+1)=\\ \frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30}$$

$$\frac{(n+1)((n+1)+1)(2(n+1)+1)(3(n+1)^2+3(n+1)-1)}{30}= \\ \frac{(n+1)(n+2)(2n+3)(3n^2+9n+5)}{30}= \\ \frac{(n^2+3n+2)(2n+3)(3n^2+9n+5)}{30}= \\ \frac{(2n^3+9n^2+13n+6)(3n^2+9n+5)}{30}= \\ \frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30}$$

You are done. You have shown that both terms are equal. Nothing has to be done backward. If you present your work you can glue the first equation block and the reverted second equation block together like this

$$\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} + (n+1)^4 =\\ \frac{n(2n^2+3n+1)(3n^2+3n-1)}{30} +(n^4+4n^3+6n^2+4n+1)=\\ \frac{n(6n^4+15n^3+10n^2 - 1)}{30} + (n^4+4n^3+6n^2+4n+1)=\\ \frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30} = \\ \frac{(2n^3+9n^2+13n+6)(3n^2+9n+5)}{30}= \\ \frac{(n^2+3n+2)(2n+3)(3n^2+9n+5)}{30}= \\ \frac{(n+1)(n+2)(2n+3)(3n^2+9n+5)}{30}= \\ \frac{(n+1)((n+1)+1)(2(n+1)+1)(3(n+1)^2+3(n+1)-1)}{30}$$

You can tell the reader that one gets the desired results by expanding both sides of $(1)$.

But there is no reason to justify the steps of the reverted second block. So you must not bother with the question

"How do I factor $$\frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30}$$ to get $$\frac{(2n^3+9n^2+13n+6)(3n^2+9n+5)}{30}$$

" nor with the question "Why do I factor $$\frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30}$$ ". Of course it is not obvious to do these steps to get the required end result (the rhs of $(1)$)

The step

$$\ldots \\ \frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30} = \\ \frac{(2n^3+9n^2+13n+6)(3n^2+9n+5)}{30} = \\ \ldots$$

is valid. This can easily proved by the reader by expanding $$\frac{(2n^3+9n^2+13n+6)(3n^2+9n+5)}{30}$$

That this step will make sense can also be seen by the reader because hewill get the desired result some lines later.

There may be aesthetical reasons to write a proof in this way but it is absolutely right to demonstrate the correctness of the equation $(1)$ by expanding both sides of the equation.

If the reader has enough practice you can simply state that $(1)$ holds and leave the proof to the reader.

Unless I missed something, this equality can't possibly hold. The leading terms on each side are different. Do you intend for this to be an identity that is generally true?

• @Bill Dubuque, is it possible for me to downgrade an answer to comment status? I'm afraid I don't know how. Still, I think I was actually answering. The answer is, "You can't do that." The LHS is not equivalent to the RHS. – MPW Jan 23 '14 at 4:34
• My apologies, I forgot to add a term at the end of the LHS. I will edit the question now. – Trent Bing Jan 23 '14 at 5:12
• It is still wrong: put $n=0$. – chubakueno Jan 23 '14 at 5:41
• @MPW The question is about how to generally prove equality of polynomials (note the remark "not asking for a worked out solution of this problem"). If there is a typo or other error in the example given then that is exactly the sort of thing that comments are designed for. You can convert to a comment by copying the answer to a comment then deleting it. – Bill Dubuque Jan 23 '14 at 5:59