Factor theorem

scrible · #1 $Old$ November 1st, 2009, 04:02 AM

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

mr fantastic · #2 $Old$ November 1st, 2009, 04:03 AM

Quote:

Originally Posted by scrible $View Post$

Hi, I having trouble with this problem $x^4-1$ 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 $(x^2)^2 - 1^2$ ....

scrible · #3 $Old$ November 1st, 2009, 03:51 PM

Quote:

Originally Posted by mr fantastic $View Post$

First note that it can be written as $(x^2)^2 - 1^2$ ....

Can it be expanded into a polynomial so that I can work with it from there?

mr fantastic · #4 $Old$ November 1st, 2009, 07:50 PM

Quote:

Originally Posted by scrible $View Post$

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 ....?

I4talent · #5 $Old$ November 2nd, 2009, 01:26 AM

Quote:

Originally Posted by scrible $View Post$

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

Method 1: $f(x) = x^4-1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mr fantastic · #6 $Old$ November 2nd, 2009, 03:57 AM

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).

Quote:

Originally Posted by I4talent $View Post$

Method 1: $f(x) = x^4-1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I4talent · #7 $Old$ November 2nd, 2009, 04:07 AM

Quote:

Originally Posted by mr fantastic $View Post$

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?

mr fantastic · #8 $Old$ November 2nd, 2009, 05:25 AM

Quote:

Originally Posted by I4talent $View Post$

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.

scrible · #9 $Old$ November 2nd, 2009, 04:10 PM

Quote:

Originally Posted by mr fantastic $View Post$

Giving hints and nudges in the right direction is usually preferable to giving a detailed and complete solution.

Thank you I got now.