Question

Please help me find all the real zeroes of the following function: g(x) = 2x^3 + x^2 - 6x - 3.

Answer 1

factor out the common factor of \(x\) first, then you will have a quadratic to solve

Answer 2

ok skip that, i didn't see the 3 at the end
it is \[ g(x) = 2x^3 + x^2 - 6x - 3\] right?

Answer 3

Correct.

Answer 4

try some easy numbers that divide 3 evenly first
1 doesn't work
how about \(-1\)?

Answer 5

nope that doesn't work either dang

Answer 6

neither does 3 or -3, on to the fractions

Answer 7

try \(x=-\frac{1}{2}\) i think that one works

Answer 8

Well, I know the steps, but I need help executing them:
1. Find all possible rational roots; the factor of the constant / the factor of the leading coefficient (p/q).
2. Test all possible roots of p/q via synthetic division.
3. Once you've found a quadratic equation, solve it for the other roots.
However, I do not know how to use p/q. May you teach me how?

Answer 9

Yes, x = -1/2 works.

Answer 10

the constant is -3
the leading coefficient is 2
the \(\frac{p}{q}\) business is fraction (or integers) where \(p\) divides the constant, and \{q\) divides the leading coefficient

Answer 11

number that go in to 3 evenly are \[\pm1,\pm3\]
that is why i suggested trying those first, because they are easiest

Answer 12

but none of those worked so now the other possible rational zeros are \[\pm\frac{1}{2}, \pm\frac{3}{2}\]

Answer 13

those are the numbers where the numerator divides 3 and the denominator divides 2

Answer 14

Okay, thank you! I finally have a better understanding of solving polynomials.

Answer 15

yw