# Thread: Help me factor a polynomial from graph thepry

1. ## Help me factor a polynomial from graph thepry

Hello, I am having a hard time understanding why k(k - 1)^n - (k - 1)^n - (-1)^n(k - 1) = (k - 1)^(n+1) + (-1)^(n+1) (k - 1).
What i specifically don't get is how to get from ^n to ^n+1. I would appreciate it if someone could tell me how this is done.

2. ## Re: Help me factor a polynomial from graph thepry

Hey takaj.

This is based on distributive law where k(k-1)^n - (k-1)^n = (k-1)^1*(k-1)^n = (k-1)^(n+1).

This is just based on ab + cb = (a+c)*b.

3. ## Re: Help me factor a polynomial from graph thepry

Originally Posted by chiro
Hey takaj.

This is based on distributive law where k(k-1)^n - (k-1)^n = (k-1)^1*(k-1)^n = (k-1)^(n+1).

This is just based on ab + cb = (a+c)*b.
Hi chiro , I don't seem to understand your explanation. Could you just go a little deeper or maybe solve it line by line. That would be great!

4. ## Re: Help me factor a polynomial from graph thepry

Let b = (k-1)^n, a = k and c = -1.

We know from distributive law that (a+c)b = ab + bc if we expand out but we can do the reverse: if we have ab + bc we can write this as (a+c)b.

Now ab = k*(k-1)^n and bc = -(k-1)^n so (a+c)b = (k + -1)*(k-1)^n = (k-1)*(k-1)^n but from powers of terms, we have z^a * z^b = z^(a+b) (exponentials) so in this case z = (k-1), a = 1 and b = n we get a+b = n+1 with our z so the final result is (k-1)^(n+1).

5. ## Re: Help me factor a polynomial from graph thepry

Originally Posted by chiro
Let b = (k-1)^n, a = k and c = -1.

We know from distributive law that (a+c)b = ab + bc if we expand out but we can do the reverse: if we have ab + bc we can write this as (a+c)b.

Now ab = k*(k-1)^n and bc = -(k-1)^n so (a+c)b = (k + -1)*(k-1)^n = (k-1)*(k-1)^n but from powers of terms, we have z^a * z^b = z^(a+b) (exponentials) so in this case z = (k-1), a = 1 and b = n we get a+b = n+1 with our z so the final result is (k-1)^(n+1).
ok, so how do you get from (-1)^n to (-1)^n+1?

6. ## Re: Help me factor a polynomial from graph thepry

Ok so your term is - (-1)^(n+1). Now this is the same as (-1)^n * (-1). But this changes your - (since this is there before the term) to a plus if you expand it out.

So your - (-1)^n * (-1) changes the - to a + giving - - (-1)^n = + (-1)^n.

7. ## Re: Help me factor a polynomial from graph thepry

Originally Posted by takaj
k(k - 1)^n - (k - 1)^n - (-1)^n(k - 1)
I found it easier this way:

Let a = k - 1, b = n - 1 ; then expression becomes:
k * a^n - a^n - a * (-1)^n

= a[k * a^b - a^b - (-1)^n]

= a[a^b * (k - 1) - (-1)^n]

= a[a^b * a - (-1)^n]

= a[a^(b + 1) - (-1)^n] ; b = n - 1 :

= a[a^n - (-1)^n]

substitute back a = k - 1

8. ## Re: Help me factor a polynomial from graph thepry

I am really not following. Could one of you recommend a website where this 'rule' is taught?

9. ## Re: Help me factor a polynomial from graph thepry

What are you not following? Stuff like a * a^(n-1) = a^n ?
If so, you need classroom help...

10. ## Re: Help me factor a polynomial from graph thepry

Originally Posted by Wilmer
What are you not following? Stuff like a * a^(n-1) = a^n ?
If so, you need classroom help...
No, no, i understand your above example.

But, tell me if this is correct...

(-1)^n *(k-1) = (-1)^(n+1) * (k-1)

11. ## Re: Help me factor a polynomial from graph thepry

This is not correct: the only reason why you can do the above is because it changes from a - to a + before the term.

12. ## Re: Help me factor a polynomial from graph thepry

Originally Posted by chiro
This is not correct: the only reason why you can do the above is because it changes from a - to a + before the term.
When the variable exponents are replaced with real numbers, I have no trouble manipulating the polynomial. But when there is a variable there, I get really confused. I don't think i need 'classroom help.'

I understand that x^n * 1^n = x^n+n, but i just don't understand how this logic is being applied to the polynomial in the original post.

actually, on rereading your original post, chiro, i think i get what you are saying. So you are manipulating the polynomial using the distribute law. I think i need to read up on this law. I know about it from logic, but i'm a bat hazy at applying it to numbers

13. ## Re: Help me factor a polynomial from graph thepry

It's because it involves -1 to some power and it's a little trick.

If you have - a*(-1)^n, then this is the same as + a*(-1)^(n+1): it's because of it being a -1 and not something else.

14. ## Re: Help me factor a polynomial from graph thepry

Originally Posted by chiro
It's because it involves -1 to some power and it's a little trick.

If you have - a*(-1)^n, then this is the same as + a*(-1)^(n+1): it's because of it being a -1 and not something else.
So i take it that a =(K-1)

That means that - (-1)^n * (k - 1) = +(-1)^n+1 * (k-1)

Am i correct?

15. ## Re: Help me factor a polynomial from graph thepry

Originally Posted by takaj
I understand that x^n * 1^n = x^n+n,...
1^(anything) = 1
So x^n * 1^n = x^n

RULE: a^n * b^n = (ab)^n

ALSO: (-1)^2n = 1 ; (-1)^(2n+1) = -1 ; where n is any integer

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