Find f(1) where f(x)=x to some obscenely large power and a few other other numbers in the polynomial, using synthetic division where the remainder is the answer.

Am I supposed to write out the same "obscenely large number" of 0 place holders, or is there an easier way to go about this (and what is it)????

Question

Find f(1) where f(x)=x to some obscenely large power and a few other other numbers in the polynomial, using synthetic division where the remainder is the answer. 

Am I supposed to write out the same "obscenely large number" of 0 place holders, or is there an easier way to go about this (and what is it)????

anonymous · Answer

add the coefficients

anonymous · Answer

Sorry?

anonymous · Answer

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$
$$f(1)=a_n+a_{n-1}+...+a_1+a_0$$

anonymous · Answer

So, when I go to solve the problem, do I have to go through each one? Sorry I'm so thick, I'm not a math person!

anonymous · Answer

$1^{n}=1$ right? so it doesn't matter what the power is

anonymous · Answer

if $f(x)=12x^{72}$ for example then $f(1)=12$

anonymous · Answer

Okay, that makes a bit more sense. If I have the problem$$f(x)=x ^{34} -7x ^{2} + 4$$ and I am trying to find the remainder from putting it in synthetic division, is the answer 31?

anonymous · Answer

no the answer is $1-7+4=-2$

anonymous · Answer

Yes, I see how that would work, but if I have to arrange the original problem for synthetic division* and follow through with it, what is the remainder?

*I'm really not sure how someone as bad as myself at verbally explaining the problems got this far in math!

anonymous · Answer

Wait! I just figured it out!

anonymous · Answer

the remainder when you divide by $x-1$ is the same as $f(1)$ which is clearly easier to compute

anonymous · Answer

Thank you for helping!

anonymous · Answer

yw