1. ## Complicated Factoring Problems

I have been going through Algebra by I.M. Gelfand and A. Shen.

"Problem 122" has been giving me difficulties. For those who have worked through this "classic" book, how does one even approach these types of problems?

My difficulties start with sub-problem "c." It wants the reader to factor a10 + a5 + 1.

In high school you only learn a handful of factoring techniques. This book shows me that is not sufficient. Where do you start?

And if you can solve this problem, could you please walk me thorough? Thank you.

2. ## Re: Complicated Factoring Problems

you have to be more explicit, not everyone owns that book.

Factor it over which domain?

3. ## Re: Complicated Factoring Problems

In fact just by observation one can say that it wont have real factors.
let us put x^5 = t the expression becomes t^2+t+1 and the discriminant for this is -3 thus it does not have real roots. If one has to go in for complex roots then use De Moivre's theorem i.e., (cos x + i sin x )^n = cos (nx) + i sin (nx) to get values for x.

4. ## Re: Complicated Factoring Problems

Originally Posted by jakncoke
you have to be more explicit, not everyone owns that book.

Factor it over which domain?
I don't know what you mean. The book only tells the reader to "factor."

After researching this more, the answer is supposed to be (a2 + a + 1)(a8 - a7 + a5 - a4 + a3 - a + 1).

How does someone get this answer? I still don't get it.

5. ## Re: Complicated Factoring Problems

Here's sub-problem "d": factor a3 + b3 +c3 - 3abc.

If I may ask here once more, how does one go about factoring something like this? Could somebody please walk me through this problem (or the above problem)?

When using this online factoring calculator, the result obtained is (c + b + a)(c2 - bc - ac + b2 - ab + a2).

6. ## Re: Complicated Factoring Problems

Hello, IdentityProblem!

I don't know how a basic Algebra text expects us to factor this polynomial.
I came up with a method, but I'm sure no one will like it (including me).

$\displaystyle \text{Factor: }\:X \:=\:a^{10} + a^5 + 1$

Multiply both sides by $\displaystyle (a^5 -1)\!:\;(a^5-1)X \:=\:(a^5-1)(a^{10}+a^5+1)$

We have: .$\displaystyle (a^5-1)X \;=\;a^{15}-1$

The left side factors:
. . $\displaystyle a^5-1 \:=\:(a-1)(a^4+a^3+a^2+a+1)$

The right side factors:
. . $\displaystyle a^{15}-1 \:=\:(a^3-1)(a^{12}+a^9+a^6+a^3+1)$
w . . . . . . $\displaystyle =\:(a-1)(a^2+a+1)(a^{12}+a^9+a^6+a^3+1)$

We have:
. . $\displaystyle (a-1)(a^4+a^3+a^2+a+1)X \;=\;(a-1)(a^2+a+1)(a^{12} + a^9 + a^6+a^3+1)$

which reduces to:
. . . . . $\displaystyle (a^4+a^3+a^2+a+1)X \;=\;(a^2+a+1)(a^{12}+a^9+a^6+a^3+1)$

Solve for $\displaystyle X$:
. . . . $\displaystyle X \;=\;\frac{(a^2+a+1)(a^{12}+a^9+a^6+a^3+1)}{a^4+a^ 3+a^2+a+1}$

Since $\displaystyle X$ is a polynomial, the 12th-degree polynomial must be divisible by the quartic.

Dividing (ugh!) we get: .$\displaystyle \frac{a^{12}+a^9+a^6+a^3+1}{a^4+a^3+a^2+a+1} \;=\; a^8 - a^7 + a^5 - a^4 + a^3 - a + 1$

Therefore: .$\displaystyle X \;=\;(a^2+a+1)(a^8-a^7+a^5-a^4+a^3-a+1)$

8. ## Re: Complicated Factoring Problems

Here is an alternative factorization.
$\displaystyle a^{10} + a^5 + 1 = 0$
$\displaystyle a^{10} + 2a^5 -a^5 +1 = 0$
$\displaystyle (a^5 + 1)^2 - (a^{5/2})^2 = 0$
$\displaystyle (a^5 + a^{5/2} + 1)(a^5 -a^{5/2} + 1) = 0$

Originally Posted by Soroban
Hello, IdentityProblem!

I don't know how a basic Algebra text expects us to factor this polynomial.
I came up with a method, but I'm sure no one will like it (including me).

Multiply both sides by $\displaystyle (a^5 -1)\!:\;(a^5-1)X \:=\a^5-1)(a^{10}+a^5+1)$

We have: .$\displaystyle (a^5-1)X \;=\;a^{15}-1$

The left side factors:
. . $\displaystyle a^5-1 \:=\a-1)(a^4+a^3+a^2+a+1)$

The right side factors:
. . $\displaystyle a^{15}-1 \:=\a^3-1)(a^{12}+a^9+a^6+a^3+1)$
w . . . . . . $\displaystyle =\a-1)(a^2+a+1)(a^{12}+a^9+a^6+a^3+1)$

We have:
. . $\displaystyle (a-1)(a^4+a^3+a^2+a+1)X \;=\;(a-1)(a^2+a+1)(a^{12} + a^9 + a^6+a^3+1)$

which reduces to:
. . . . . $\displaystyle (a^4+a^3+a^2+a+1)X \;=\;(a^2+a+1)(a^{12}+a^9+a^6+a^3+1)$

Solve for $\displaystyle X$:
. . . . $\displaystyle X \;=\;\frac{(a^2+a+1)(a^{12}+a^9+a^6+a^3+1)}{a^4+a^ 3+a^2+a+1}$

Since $\displaystyle X$ is a polynomial, the 12th-degree polynomial must be divisible by the quartic.

Dividing (ugh!) we get: .$\displaystyle \frac{a^{12}+a^9+a^6+a^3+1}{a^4+a^3+a^2+a+1} \;=\; a^8 - a^7 + a^5 - a^4 + a^3 - a + 1$

Therefore: .$\displaystyle X \;=\;(a^2+a+1)(a^8-a^7+a^5-a^4+a^3-a+1)$

9. ## Re: Complicated Factoring Problems

To tackle problem "d" this, first remember that

1. $\displaystyle (a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$
2. (1) can be rewritten as $\displaystyle (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
3. (2) allows us to obtain $\displaystyle a^3 + b^3 = (a + b)^3 - 3ab(a + b)$
4. We need to factor $\displaystyle a^3 + b^3 + c^3 - 3abc = 0$
5. Grouping the first 2 terms in (4) $\displaystyle (a^3 + b^3) + c^3 - 3abc = 0$
6. Using (3), (5) can be rewritten as $\displaystyle (a + b)^3 + c^3 - 3ab(a+b) -3abc$
7. Taking -3ab common from the last 2 terms in (6) we get $\displaystyle (a + b)^3 + c^3 - 3ab(a + b + c)$
8. The first 2 terms in 7 can be given the same treatment as in (5) and then we get $\displaystyle (a + b+ c)^3 - 3c(a + b)(a + b + c) - 3ab(a + b+ c)$
9. Rewriting 8 .. $\displaystyle (a + b + c)((a+ b+ c)^2 - 3ac - 3bc - 3ab)$
10. Expanding $\displaystyle (a + b+ c)^2$ in 9 causes it to reduce to the simpler final form $\displaystyle (a + b + c)(a^2 + b^2 + c^2 -ab -ac -bc)$

Originally Posted by IdentityProblem
Here's sub-problem "d": factor a3 + b3 +c3 - 3abc.

If I may ask here once more, how does one go about factoring something like this? Could somebody please walk me through this problem (or the above problem)?

When using this online factoring calculator, the result obtained is (c + b + a)(c2 - bc - ac + b2 - ab + a2).

10. ## Re: Complicated Factoring Problems

Thank you all.

This does help. I'll have to go over these two problems a couple of times before it sinks in, though. Hopefully after that the next couple of problems I'll get myself.

And it is indeed beyond "basic" algebra. (Maybe I should have posted it in the Pre-Calculus Help Forum.)