# Factor theorem proof: Why don't we write p(x) - p(y) as...?

#### Aion

##### New member
Since

$$\displaystyle \frac{x^k - y^k}{x-y} = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1}$$

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

hence $$\displaystyle {x^k - y^k} = (x-y)q_k(x)$$

As usual, we write our polynomial $$\displaystyle p$$ as

$$\displaystyle p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$$

$$\displaystyle p(x)-p(y) =$$

$$\displaystyle a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$$

$$\displaystyle \\ -(a_ny^n + a_{n-1}y^{n-1} + ... + a_1y + a_0)$$

$$\displaystyle = a_n(x^n-y^n) + a_{n-1}(x^{n-1}-y{n-1}^) + ... + a_1(x-y)$$

The terms have the form $$\displaystyle a_k(x^k-y^k)$$. But $$\displaystyle x^k-y^k = (x-y)q_k(x)$$, and if we substitute this in we get:

$$\displaystyle p(x)-p(y)=$$

$$\displaystyle a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)$$

$$\displaystyle (x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1)$$

$$\displaystyle = (x-y)q(x)$$ where $$\displaystyle q(x) = a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1$$

If $$\displaystyle p(a) = 0$$ where $$\displaystyle y = a$$ then

$$\displaystyle p(x)-p(a) = (x-a)q(x)$$

Q.E.D

My question is why we don't write $$\displaystyle p(x) - p(y)$$ as

$$\displaystyle a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)q_1(x)$$

$$\displaystyle (x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1q_1(x))$$.

Can anyone please explain this to me thanks!

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#### Dr.Peterson

##### Elite Member
Since

$$\displaystyle \frac{x^k - y^k}{x-y} = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1}$$

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

hence $$\displaystyle {x^k - y^k} = (x-y)q_k(x)$$

As usual, we write our polynomial $$\displaystyle p$$ as

$$\displaystyle p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$$

$$\displaystyle p(x)-p(y) =$$

$$\displaystyle a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$$

$$\displaystyle \\ -(a_ny^n + a_{n-1}y^{n-1} + ... + a_1y + a_0)$$

$$\displaystyle = a_n(x^n-y^n) + a_{n-1}(x^{n-1}-y^{n-1}) + ... + a_1(x-y)$$ [I fixed this for you]

The terms have the form $$\displaystyle a_k(x^k-y^k)$$. But $$\displaystyle x^k-y^k = (x-y)q_k(x)$$, and if we substitute this in we get:

$$\displaystyle p(x)-p(y)=$$

$$\displaystyle a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)$$

$$\displaystyle (x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1)$$

$$\displaystyle = (x-y)q(x)$$ where $$\displaystyle q(x) = a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1$$

If $$\displaystyle p(a) = 0$$ where $$\displaystyle y = a$$ then

$$\displaystyle p(x)-p(a) = (x-a)q(x)$$

Q.E.D

My question is why we don't write $$\displaystyle p(x) - p(y)$$ as

$$\displaystyle a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)q_1(x)$$

$$\displaystyle (x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1q_1(x))$$.

Can anyone please explain this to me thanks!

Interesting proof; now at last I can see why you previously perceived circularity in the use of the fact it starts with.

By the way, before showing a proof, you should always show the statement of the theorem, even if you think everyone knows what the theorem name refers to; it may not always use the same names.

Clearly they think that q1(x) = 1, so that they didn't need to write q1(x) there. So your task in reading this is to check that implicit claim. (If the proof was intended for beginners, it would be kind to state this explicitly, and perhaps even prove it, but proofs commonly assume that the reader can fill in gaps up to some size.)

So let's look. What would q1 be? They didn't explicitly define qk(x), but implicitly give it as

$$\displaystyle q_k(x) = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1}$$

Therefore, letting k = 1,

$$\displaystyle q_1(x) = x^{1-1}y^0 = x^0y^0 = 1$$

So they are right. Rather than write q1(x), they can just drop it.

Similarly, they could have written

$$\displaystyle q_k(x) = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + x^0y^{k-1}$$

but didn't happen to show x0, though they did show y0. That's just a little inconsistency in their choices.

Got it?

#### jonlara3

##### New member
Thank you so much for kin your information,,I hope any other people benefited.