Math Prob 3?

Question

Solve the following equation:

(1/3)x^3 - (1/2)x^2 + (1/3)x + (1/3) = 0

Answer 1

Use synthetic division, email if you need help. Don't just ask for the answer, you need to understand the concept. I know, I failed my first test (i've always maintained A's and occasional B's), recently because of this kind of studying.

Answer 2

(1/3)x^3 - (1/2)x^2 + (1/3)x + (1/3) = 0

Get rid of fractions by multiplying through by 6 :
2x^3 - 3x^2 + 2x + 2 = 0

There are various ways to do this, but probably the
easiest here, is to use the Rational Root Theorem,
which states, in relation to your equation :

If a 1st degree polynomial, ax + b, with integer coefficients,
is a factor of another polynomial, p(x) = cx^2 + dx + e,
with integer coefficients, then a is a factor of c,
while b is a factor of e.

So the equation (if it has a rational root) can be represented as :
(ax + b)(cx^2 + dx + e) = 0

Expanding this and simplifying gives :
(ac)x^3 + (ad + bc)x^2 + (ae + bd)x + be = 0

Correlating this with the equation :
2x^3 - 3x^2 + 2x + 2 = 0, gives :

ac = 2 and be = 2

We're interested in a and b from (ax + b),
which is to be, the (hopefully), rational factor.

From ac = 2 and be = 2, the only integers
that a and b can be, are : ± 1 and ± 2.

So (ax - b) can only be : (± x ± 1), (± x ± 2), (± 2x ± 1), (± 2x ± 2)

We can dispense with the -x solutions, because
the coefficient of x^3 in the equation is positive.
And we can ignore (± 2x ± 2), because it gives the
same solutions as (± x ± 1), as we can factor out the 2.

This leaves just : (x ± 1), (x ± 2) and (2x ± 1)
as being the only 6 possible factors.

This means that x could be ± 1, ± 2 or ± 1/2.

Substituting each one of these 6 possibilities into
y = 2x^3 - 3x^2 + 2x + 2, gives :

x = -1 implies y = -5
x = 1 implies y = 3
x = -2 implies y = -30
x = 2 implies y = 10
x = -1/2 implies y = 0
x = 1/2 implies y = 5/2

So we have a root, because y = 0 when x = -1/2.

x = -1/2 comes from (2x + 1), so this is a factor.

Dividing (2x + 1) into (2x^3 - 3x^2 + 2x + 2)
by synthetic division yields :

2x^3 - 3x^2 + 2x + 2 = (2x + 1)(x^2 - 2x + 2) = 0

Thus, x = -1/2 and the two complex conjugate roots,
(1 ± i), which can be obtained by applying the
quadratic formula to x^2 - 2x + 2 = 0.