Question

Find the roots of each polynomial function.

Answer 1

\[x ^{3}+2x ^{2}+3x+6=0\]

\[8x ^{4}-66x ^{3}+175^{2}-132x-45=0\]

Answer 2

just gotta factor those puppies

Answer 3

I don't know how to do that.. @TheSeabass

Answer 4

do you know anything about factoring first of all?

Answer 5

to an extent, I don't really understand it

Answer 6

well i was hoping you did cuz ive forgotten exactly how to do these

Answer 7

lol, it's all good. thanks anyway :)

Answer 8

Yeah sorry about that mabe try asking one of the mods, you know, the ones thats show when you click the members thing on the top screen. the ones with fancy colors

Answer 9

To factor \[x^3+2x^2+3x+6=0\]try grouping the terms by pairs and factoring each pair:
\[(x^3+2x^2) + (3x+6) = 0\]\[x^2(x+2) + 3(x+2) = 0\]Now both have \((x+2)\) as a common factor, so factor that out, giving you \[(x+2)(x^2+3) = 0\]Use the zero product property to find the roots from that...

For the other one, you might use the rational root theorem to guess at a root. Then try it out in the polynomial and see if it gives you 0. If it does, it is a root, and you can divide the polynomial by \((x-root)\) to get a simpler polynomial with the same remaining roots. You may be able to factor that polynomial as I did above, or guess another root, etc.

Answer 10

how do I use the rational root theorem? @whpalmer4

Answer 11

Ah, sorry, I missed the notification that you'd replied.

The RRT makes use of the fact that the constant term (-45 for your second question) is formed by multiplying the coefficients of the leading and trailing terms. It's simpler to grasp if the highest exponent term has a coefficient of 1: there you simply have \[(x+a)(x+b)(x+c) = x^3 + ... + abc\]for example. You take the constant term, factor it, and the positive and negative factors are the possible rational roots. Unfortunately, you end up with more possible roots than actual roots, but it's a start.

Now if term with the largest exponent has a coefficient other than 1 (such as in this case), things get a bit more complicated. In that case, assuming I'm remembering correctly, you have as potential roots the positive and negative values of all of the combinations of factors of the constant term as a numerator and all of the factors of the coefficient of the highest exponent term as the denominator. Here, that means

\[\pm\frac{1,3,5,15,45}{1,2,4,8}\]

I think that's right, though if you find it confusing, I do, too :-)

A hint: try the numbers which aren't fractions first.

Another hint: you can use "synthetic substitution" which is just like synthetic division. If you get a 0 at the end, it means the number you were trying is a root.