1. ## Factor theorem

Hi, I having trouble with this problem $\displaystyle x^4-1$ it saids to factor this problem as much as possible. Can some one please help me with this problem?

2. Originally Posted by scrible
Hi, I having trouble with this problem $\displaystyle x^4-1$ it saids to factor this problem as much as possible. Can some one please help me with this problem?
First note that it can be written as $\displaystyle (x^2)^2 - 1^2$ ....

3. Originally Posted by mr fantastic
First note that it can be written as $\displaystyle (x^2)^2 - 1^2$ ....
Can it be expanded into a polynomial so that I can work with it from there?

4. Originally Posted by scrible
Can it be expanded into a polynomial so that I can work with it from there?
Have you been taught how to factorise the difference of two squares ....?

5. Originally Posted by scrible
Hi, I having trouble with this problem $\displaystyle x^4-1$ it saids to factor this problem as much as possible. Can some one please help me with this problem?
Method 1: $\displaystyle f(x) = x^4-1$

$\displaystyle f(1) = (1)^4-1 = 1-1 = 0$, therefore $\displaystyle (x-1)$ is a factor of $\displaystyle x^4-1$.
$\displaystyle f(-1) = (-1)^4-1= 1-1 = 0$, therefore $\displaystyle (x+1)$ is a factor of $\displaystyle x^4-1$.

So we have $\displaystyle x^4-1 = (x-1)(x+1)(ax^2+bx+c)$
$\displaystyle (x^2-1)(ax^2+bx+c) = x^4-1$
$\displaystyle x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1$
$\displaystyle ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1$
$\displaystyle (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1$

$\displaystyle \begin{cases} a = 1 \\ b = 0 \\ c-a = 0 & c-1 = 0 \implies c = 1 \\ -b = 0 \\ -c = -1 & c = 1. \end{cases}$

Therefore $\displaystyle ax^2+bx+c = x^2+1$

Hence $\displaystyle \boxed{(x^2-1)(x^2+1) = x^4-1}$

Method 2: Any products yielding answers of the form $\displaystyle a^n-b^n$ are given by:

$\displaystyle (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-1}+b^{n-1}) = a^n-b^n$

So for $\displaystyle x^4-1^2$, we have:

$\displaystyle (x-1)((x^{4-1})+(x^{4-2})({1})+(x^{4-3})({1})^2+(x^{4-4})({1})^3)= x^4-1^2$

$\displaystyle (x-1)((x^{3})+(x^{2})({1})+(x^{1})({1})^2+(x^{0})({1} )^3)= x^4-1^2$

$\displaystyle (x-1)(x^3+x^2+x+1) = x^4-1$

$\displaystyle x^3+x^2+x+1$ yields to zero when $\displaystyle x = -1$, so, by the factor theorem:

$\displaystyle (x+1)(x-1)(ax^2+bx+c) = x^4-1$

$\displaystyle (x^2-1)(ax^2+bx+c) = x^4-1$
$\displaystyle x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1$
$\displaystyle ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1$
$\displaystyle (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1$

$\displaystyle \begin{cases} a = 1 \\ b = 0 \\ c-a = 0 & c-1 = 0 \implies c = 1 \\ -b = 0 \\ -c = -1 & c = 1. \end{cases}$

Therefore $\displaystyle ax^2+bx+c = x^2+1$

Hence $\displaystyle \boxed{(x^2-1)(x^2+1) = x^4-1}$

6. I appreciate the time and effort you have put into this reply. However, when someone gives hints on how to solve a problem, please do not then give a full solution; it is forum etiquette to wait until the original poster has responded before replying. (And I suspect that the path I'm trying to steer the OP down is a more accessible one).

Originally Posted by I4talent
Method 1: $\displaystyle f(x) = x^4-1$

$\displaystyle f(1) = (1)^4-1 = 1-1 = 0$, therefore $\displaystyle (x-1)$ is a factor of $\displaystyle x^4-1$.

$\displaystyle f(-1) = (-1)^4-1= 1-1 = 0$, therefore $\displaystyle (x+1)$ is a factor of $\displaystyle x^4-1$.

So we have $\displaystyle x^4-1 = (x-1)(x+1)(ax^2+bx+c)$

$\displaystyle (x^2-1)(ax^2+bx+c) = x^4-1$

$\displaystyle x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1$

$\displaystyle ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1$

$\displaystyle (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1$

$\displaystyle \begin{cases} a = 1 \\ b = 0 \\ c-a = 0 & c-1 = 0 \implies c = 1 \\ -b = 0 \\ -c = -1 & c = 1. \end{cases}$

Therefore $\displaystyle ax^2+bx+c = x^2+1$

Hence $\displaystyle \boxed{(x^2-1)(x^2+1) = x^4-1}$

Method 2: Any products yielding answers of the form $\displaystyle a^n-b^n$ are given by:

$\displaystyle (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-1}+b^{n-1}) = a^n-b^n$

So for $\displaystyle x^4-1^2$, we have:

$\displaystyle (x-1)((x^{4-1})+(x^{4-2})({1})+(x^{4-3})({1})^2+(x^{4-4})({1})^3)= x^4-1^2$

$\displaystyle (x-1)((x^{3})+(x^{2})({1})+(x^{1})({1})^2+(x^{0})({1} )^3)= x^4-1^2$

$\displaystyle (x-1)(x^3+x^2+x+1) = x^4-1$

$\displaystyle x^3+x^2+x+1$ yields to zero when $\displaystyle x = -1$, so, by the factor theorem:

$\displaystyle (x+1)(x-1)(ax^2+bx+c) = x^4-1$

$\displaystyle (x^2-1)(ax^2+bx+c) = x^4-1$

$\displaystyle x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1$

$\displaystyle ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1$

$\displaystyle (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1$

$\displaystyle \begin{cases} a = 1 \\ b = 0 \\ c-a = 0 & c-1 = 0 \implies c = 1 \\ -b = 0 \\ -c = -1 & c = 1. \end{cases}$

Therefore $\displaystyle ax^2+bx+c = x^2+1$

Hence $\displaystyle \boxed{(x^2-1)(x^2+1) = x^4-1}$

7. Originally Posted by mr fantastic
I appreciate the time and effort you have put into this reply. However, when someone gives hints on how to solve a problem, please do not then give a full solution; it is forum etiquette to wait until the original poster has responded before replying. (And I suspect that the path I'm trying to steer the OP down is a more accessible one).
I'm sorry. I was not aware of that. But does this rule - or convention - apply only when someone has given a hint, or does everybody have to give hints and not the whole solution unless it's the last resort?

8. Originally Posted by I4talent
I'm sorry. I was not aware of that. But does this rule - or convention - apply only when someone has given a hint, or does everybody have to give hints and not the whole solution unless it's the last resort?
Giving hints and nudges in the right direction is usually preferable to giving a detailed and complete solution.

9. Originally Posted by mr fantastic
Giving hints and nudges in the right direction is usually preferable to giving a detailed and complete solution.
Thank you I got now.