On Wed, 24 Oct 2007 16:21:04 -0400, Greg Neill
<gneillREM@[EMAIL PROTECTED]
> wrote in
<news:471fa91e$0$22111$9a6e19ea@[EMAIL PROTECTED]
> in
alt.algebra.help:
> "Dougsd1r" <dougsdir@[EMAIL PROTECTED]
> wrote in message
> news:1193230117.800631.320390@[EMAIL PROTECTED]
>> How do I fully solve this equation?
>> x^3 + x^2 -5x + 3 = 0
>> According to MS Math there are three solutions x=-3, x=1
>> and x=1
>> here is what i have so far:
>> x^3 + x^2 -5x = -3
>> (x)(x^2 + x - 5) = -3
>> now im stuck :( any helpers?
> Leave the right hand side (of the original) alone
> and factor the left hand side. With a polynomial,
> you can sometimes find at least one root by just
> looking at the coefficients. Here the coefficient
> of the leading term (x^3) is 1 and the constant
> coefficient is 3, so try +/- 1*3, that is, +3 and
> -3 in the equation.
You don't want to multiply the coefficients; in general that
doesn't work. What you want is the rational root test:
Let p(x) = a_0 + a_1*x + a_2*x^2 + ... + a_n*x^n
be a polynomial with integer coefficients, where
a_n != 0. (In other words, the polynomial really
is of degree n.) If a rational numbers r/s is a root
of p(x), then r must be a divisor of a_0, and s must
be a divisor of a_n. That is, the numerator of the
rational root must be a divisor of the constant
term of the polynomial, and the denominator of
the rational root must be a divisor of the leading
coefficient of the polynomial.
Note that there is no guarantee that any of these rational
numbers actually *is* a root of p(x); the test says only
that these rational numbers are the only rational numbers
that could possibly be roots of p(x).
In the original problem we want a root of the polynomial
p(x) = x^3 + x^2 - 5x + 3. Suppose that r/s is a solution
of p(x) = 0, where r and s are integers, and the fraction is
in lowest terms. The rational root test says that r must be
a divisor of the constant term 3, and s must be a divisor of
the leading coefficient 1. Thus, r must be 1, -1, 3, or -3,
and s must be 1 or -1. The possible quotients are 1/1 =
-1/-1 = 1, -1/1 = 1/-1 = -1, 3/1 = -3/-1 = 3, and -3/1 =
3/-1 = -3.
Start with the easiest, and try x = 1:
p(1) = 1^3 + 1^2 - 5*1 + 3 = 0,
so 1 is a root of the polynomial, and therefore x - 1 is a
factor of p(x). (General fact, in case you've forgotten: r
is a root of a polynomial p(x) if and only if x - r is a
factor of p(x).) Use ordinary polynomial long division to
divide p(x) by x - 1 and get a quotient of x^2 + 2x - 3.
You can then factor this by inspection to complete the
factorization of the original cubic.
Brian
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