Thread: Cubics fun

There are two parts for this question.

1. show that, for any constant a and any natural number n, x - a is a factor of x^n - a^n.
2. find conditions(if any) on n that are required in order that:
i. x + a is a factor of x^n - a^n
ii. x + a is a factor of x^n - a^n

this is probably a big ask but I really really want to understand the theory behind this so I really need to get this properly explained!

thanks in advance!!

Originally Posted by delicate_tears

There are two parts for this question.

1. show that, for any constant a and any natural number n, x - a is a factor of x^n - a^n.

Consider:



Multiply this by to get:



Now instead multiply it by :



Now if we subtract these last two expressions all but the first term of the first and the last term of the second cancell out, so:

.

Which proves that is a factor of

(Please clarrify question 2 as the two parts are the same at present).

RonL

Originally Posted by delicate_tears

There are two parts for this question.

1. show that, for any constant a and any natural number n, x - a is a factor of x^n - a^n.
2. find conditions(if any) on n that are required in order that:
i. x + a is a factor of x^n - a^n
ii. x + a is a factor of x^n - a^n Mr F queries: Do you mean x^n + a^n?

this is probably a big ask but I really really want to understand the theory behind this so I really need to get this properly explained!

thanks in advance!!

Alternatively to the good Captain:

1. Let p(x) = x^n - a^n. Note that p(a) = 0. By the factor theorem it follows that x - a is a factor.

2. i. x + a will only be a factor if p(-a) = 0 .....

Originally Posted by mr fantastic

Alternatively to the good Captain:

1. Let p(x) = x^n - a^n. Note that p(a) = 0. By the factor theorem it follows that x - a is a factor.

2. i. x + a will only be a factor if p(-a) = 0 .....

I must say that I assumed that the student had not yet done the factor theorem (since otherwise this all seems a bit too easy) - but then what do I know?

Also the second part of the question becomes quite interesting without the factor theorem.

RonL

Originally Posted by CaptainBlack

I must say that I assumed that the student had not yet done the factor theorem (since otherwise this all seems a bit too easy) - but then what do I know?

[snip]

Well, you know what they say .... Never underestimate how difficult something easy can be.

Originally Posted by CaptainBlack

[snip]

Also the second part of the question becomes quite interesting without the factor theorem.

RonL

Agreed.