Math help- polynomials (1 Viewer)

[B1andB2]

Hey guys, for a question likes this,
P(x) = x^4 − 6x^3 − 19x^2 + 84x + 180. where the constant is really large and it's a quartic, is there a shorter way to do it, other than finding a factor, long diving and repeating the process multiple times? Is there a way i can someone decrease the 180?
Thanks

 

[Minari243]

Search up a method called synthetic division, its basically a faster and less error prone method of long division for polynomials

 

[Velocifire]

Yeh, saved me in a test.

But is there no choice other than the boring long division if the leading coefficient of the poly is non-monic? Like if that x^4 was replaced by 2x^4

 

[idkkdi]

Hey guys, for a question likes this,
P(x) = x^4 − 6x^3 − 19x^2 + 84x + 180. where the constant is really large and it's a quartic, is there a shorter way to do it, other than finding a factor, long diving and repeating the process multiple times? Is there a way i can someone decrease the 180?
Thanks

Sum/product of roots. Differentiating to see if it is multiple multiplicity. Slap on the best technique.

 

[B1andB2]

Sum/product of roots. Differentiating to see if it is multiple multiplicity. Slap on the best technique.

That's some big brain thing right there LOL

 

[idkkdi]

That's some big brain thing right there LOL

Nice sarcasm?
Your teacher won't ask you a quartic like that which can't be factorised first, or without a given/easy root, or without a condition of a root being multiple multiplicity unless they are retarded.

 

[Velocifire]

Was thinking that too, since there's a shortcut for that for sum of the roots in a quartic and for products of roots.

(given p(x) = ax^4 bx^3 +cx^2 +dx +e)

 

[idkkdi]

hey is there a shorctut for that? apparently theres like a way to easily find out a+b+g or like abg

Cambridge MX1 Chp 10, F, I believe, or maybe E.

 

[B1andB2]

Nice sarcasm?
Your teacher won't ask you a quartic like that which can't be factorised first, or without a given root, or without a condition of a root being multiple multiplicity unless they are retarded.

The question is from MIF. I did it without long division or differentiation, all g!

 

[Velocifire]

so you used synthetic?

 

[idkkdi]

The question is from MIF. I did it without long division or differentiation, all g!

???????????????????????? how

 

[Velocifire]

Synthetic Division, anyone can do that!

Easier that LD and definitely easier than Differentiation and provided that P(x) is non-monic, it's pretty easy.

 

[B1andB2]

???????????????????????? how

so you used synthetic?

I did this... *also i am aware that i skip some steps in my working out for the remainder theorem pls don't call me out lol

 

[idkkdi]

I did this... *also i am aware that i skip some steps in my working out for the remainder theorem pls don't call me out lol

Oh, those factors are easy to find, that's fair.

 

[B1andB2]

Oh, those factors are easy to find, that's fair.

Hey do you bother doing qs from MIF? Or do you just use cambridge?

 

[idkkdi]

Hey do you bother doing qs from MIF? Or do you just use cambridge?

School doesn't use MIF, it's easier? Most top 10, maybe all use Cambridge afaik.

 

[foxxraven]

I did this... *also i am aware that i skip some steps in my working out for the remainder theorem pls don't call me out lol

holy crap, your handwriting is gorgeous.

 

[B1andB2]

holy crap, your handwriting is gorgeous.

Thank u

 

[Velocifire]

Yeah, was thinking that as well, better than mine, pretty hard to confine letters to one square if using a grid book (for me)

 

[B1andB2]

Just in case someone sees this thread and hasn't heard of my method:
P(x) = x^4 − 6x^3 − 19x^2 + 84x + 180
180 is pretty big so I plug in integers which I can quickly see are divisible by 180.
The constant is there so I don't start with 0. I plugged in 1 in the calculator, and = (to set Ans to 1) then cleared and replaced the x with the Ans key in P(x) and see if you get 0. I did the same with 2, 3, 4, etc and then when you get big or small enough, do the other side. -1, -2, -3, etc.
Pretty easy for me to find the roots are 5,6,-2,-3.
and yes B1andB2's working out is correct, and definitely required in an exam. But for practice, you can skip those steps XD

Thank u!