Question

List all possible zeroes for the polynomial:
h(x)=2x^3+5x^2-31x-15

Answer 1

Is the answer:+-1, +-2, +-3, +-3/2, +-5, +-5/2, +-15, +-15/2

Answer 2

@jdoe0001

Answer 3

have you covered the "rational root test" yet?

Answer 4

yes, that's how I got this answer

Answer 5

yes, with the exception of \(\pm 2\) because recall that "2" will be the denominator part, not the numerator, so it'd be more like \(\bf \pm \cfrac{1}{2}\)

Answer 6

Oh right I forgot. So using this equation I need to use descartes rule of signs to describe the possible combinations of roots

Answer 7

f(x) gives the possible POSITIVE real roots
h(x)= +2x^3 +5x^2 -31x -15
^ ^ ^
no yes no

signs changed once, so real positive ones are 1 or 0

f( -x ) gives the possible NEGATIVE real roots
h( -x )= -2x^3 +5x^2 +31x -15
^ ^ ^
yes no yes

signs changed twice, so real negative ones are 2, or 0

so, it has , 1 positive and 2 negative ones

Answer 8

So I would write one real positive roots and two negative real roots? It makes a lot more sense now

Answer 9

yes

Answer 10

nice!