Precalc bonuses

hey /sci/, first time poster, here's an interesting set of questions

not my homework, just some bonuses on my last precalculus test i was intrigued by.

first is (oh and x^y means x to the y power);
what do you get when you divide x^74-5x^60+x^23+x^6 by x-2?

this one i'm stumped on, you would have to write out all the intermediate x values with a zero coefficient, right?


last one was: write a polynomial with the roots (4, 5, i2). I know the 4 and 5 would be (x-5) and (x-4), but what about that i2? how would you work BACKWARDS through the quadratic formula?

1. Polynomial division. Use it. Also zero coefficient = nothing at all. Idiot.
2. (x-5)(x-4)(x-2i)(x+2i)
Quadratic formula backwards? WTF are you talking about?

lern2math

The first one seems needlessly lengthy, unless there's some clever way of doing it I can't see.

x^74-5x^60+x^23+x^6 =
x^6(x^68-5x^54+x^17+1) =
x^6(x^54(x^14-5)+x^17+1)...

I see no obvious way to further factor that monster.  2^14-5 >> 0, so 2 is not a root of the original polynomial -> there will be a remainder when divided by x-2.  I guess you're chugging out long division for the next hour to find that remainder.  There exists an algorithm for efficient long division, but even that will take a while.

>>2 nicely answered your 2nd concern.  Backwards through quadratic formula = multiply out the factors?

turns out that the ACTUAL question is (i got the test back)

give the REMAINDER when P(x)=x^72-5x^24+8x^13+2x^7-x+7 is divided by x-1.

i completely forgot about the remainder part, this can EASILY be done by the remainder theorem, which basically states to throw a 1 (from the x-1) in there and report the result.

d'oh

cring out loud

Precalc boners?

In calc  throw in a 1 and get your answers.