Question

I hate finding real roots of cubic polynomials. Any tricks?

Answer 1

1) use wolfram alpha it will tell you
2) use a graphing calculator and guess from there
3) if the constant is
\[a_0\] and the leading coefficient is
\[a_3\] then the possible "rational roots"
\[\frac{p}{q}\] must have p divides
\[a_0\] and q divides \[a_3\]

Answer 2

besides calculator or wolfram...

Answer 3

and also if the problem is from a textbook or your teacher, look for the obvious ones first. like 1 and -1

Answer 4

Ok, so refresh my memory here satellite (since I haven't done this in like 3 years). . . integer factors of the constant term are all possibly real roots, right?

Answer 5

for example if you see
\[p(x)=x^3-3x^2+x-3\] only possible zeros are
\[\pm1,\pm3\]

Answer 6

i mean only possible RATIONAL zeros.

Answer 7

oh right, so the real roots can be rational, not just integers

Answer 8

of course the zeros could be
\[1+\sqrt{3}\] and \[1-\sqrt{3}\] for example. no one says they have to be rational. or even real

Answer 9

real roots could be integers, fractions or irrational.

Answer 10

but if you have a cubic polynomial it must have one real zero

Answer 11

I'm just looking for eigenvalues for 3x3 matrices, so I need to be able to find something to divide by synthetic division-wise to get other roots with cubic poly

Answer 12

Cause I just KNOW dr. y is gonna have one of these on the exam tomorrow

Answer 13

well if you live right one of the zeros will be obvious and once you have one you can factor and find the other two via quadratic formula

Answer 14

right..that's my plan..just going through textbook and couldn't remember the "rule" on how to figure out what my options were to divide by

Answer 15

yeah but he is a nice guy so one of the zeros will be 1 or -1 or 2 or something easy

Answer 16

I just need to be careful expanding and collecting like terms. i'm making mistakes like I'm in 017 or something.

Answer 17

i mean it can be annoying if you have
\[p(x)=2x^3-5x+6\] then you have all sorts of possible rational zeros.
\[\pm1,\pm2,\pm3\pm6,\pm\frac{1}{2},\pm\frac{3}{2}\]

Answer 18

look for easy ones first is my best guess. why can't you use a calculator?

Answer 19

Oh, I remember. The factors of the constant divided by the lead coefficient are also possibilities.