Use long division to perform the division (Express your answer as quotient + remainder/divisor.)
\[x^4+8x^3-4x^2+x-2 \over x-2\]
do you know how to use 'synthetic division'? i can try to write it if you do not
i am no good with division at all
oh actually it says "long division" doesn't it. ok then take out paper and pencil and write like you would a regular long division problem \[x-2|x^4+8x^3-4x^2+x-2\]
something like that. or would you just like to use synthetic division it is much much easier.
i believe it wants me to work it out the way you wrote it up top
list the coefficients of the numerator 1 8 -4 1 -2
ok they we are in for a world of annoyance. fine, write what i did first. now forget about the -4 for a moment. what is \[x^4\] divided by \[x\]?
in other words how many times does \[x\] go in to \[x^4\] or even more simply what is \[\frac{x^4}{x}\]?
1?
no, try this. what is \[\frac{2^4}{2}\]?
8
yes, aka \[2^3\]
what do you think \[\frac{5^4}{5}\] is without computing
what do you mean without computing?
i mean write your answer as 5 to a power
not sure i follow...
ok lets try this \[\frac{x^4}{x}=\frac{x\times x \times x\times x}{x}\] and when you cancel one of the x's what do you get?
you should get x to a power yes?
Or you can think of it as "What do I need to multiply times x to get \(x^4\)"
polpak you get the award for the day with \[b^0\]!
=)
now lets see if we (you) can help mathater realize that \[\frac{x^4}{x}=x^?\]
because we have a long division to do.
I don't think it'll work cause they're not here anymore ;)
yes, the problem is variables are confusing if you are not used to them. that is why \[\frac{x^4}{x}=1\] cancel the x's and \[1^4=1\]!
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