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Factor: x^3 - 3/4x - 1/4

The answer is: (x - 1)(x + 1/2)^2

How do I learn to do that? I'd like to reread an appropriate chapter from an appropriate textbook and do practice problems.

It takes experience and practiced eye. Algebra books have chapters on factoring; ask your teacher to borrow his/her desk copies. Also search google, such as library thinkquest for material.

You first multiply the equation by 4 so that all the coefficients are integers:

4x^3 - 3x - 1

You then reason as follws. Suppose that this function has a rational root. I.e. there exists integers p and q that are reletively prime (i.e. they don't have any prime divisors in common) such that p/q is a root. Then it can be shown (I'll give the proof below) that:

p must divide -1

q must divide 4

So all the candidates for rational roots are:

+/- 1, +/- 1/2 and +/-1/4

If yopu find one root, say x = 1, then you just divide 4x^3 - 3x - 1 by x - 1 using e.g. long division and you then find a quadratic equation that you can factor using elementary means and you'll obtain the factor (x + 1/2)^2


Suppose you have an equation of the form

a_{n} x^n + a_{n-1} x^(n-1) + ...a_{1}x + a_{0} = 0

where all the a_{j} are integers. Suppose that a solution is x = p/q where p and q are integers that are relatively prime. Inserting x = p/q and multiplying by q^n gives:

a_{n} p^n + a_{n-1} p^(n-1)q + ...a_{1}pq^(n-1) + a_{0}q^n = 0

Let's write this as:

a_{n} p^n + a_{n-1} p^(n-1)q + ...a_{1}pq^(n-1) =- a_{0}q^n

Note that the left hand side is divisible by p:

[a_{n} p^(n-1) + a_{n-1} p^(n-2)q + ...a_{1}q^(n-1)]p = =- a_{0}q^n

The square brackets on the left hand side is an integer.

This then means that the right hand side must also be divisible by p. But q has no factors in common with p, so a_{0} must be divisible by p!

You can also easily see that a_{0} must be divisible by q by rewriting the equation as:

a_{n-1} p^(n-1)q + ...a_{1}pq^(n-1) + a_{0}q^n = -a_{n} p^n

You now observe that the left hand side is divisible by q, therefore the right hand side must also be divisible by q. Because p has no factors in common with q you can conclude that a_{n} must be divisible by q.


"You can also easily see that a_{0} must be divisible by q by rewriting the equation as:"

I meant a_{n}, of course :)

that's very helpful. Thank you so much for the assistance!

Count Iblis, that was presented to the student in a most informative and helpful way. Thanks.

Thanks! It's a pleasure to help here!

You are welcome :)

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