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### Factoring Quartics

```
Date: 10/20/96 at 15:54:24
From: Anonymous
Subject: Dr Math

Can you help me factorize f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1
and solve f(x) = 0?

Thank you in advance.
```

```
Date: 10/22/96 at 19:2:15
From: Doctor Yiu
Subject: Re: Dr Math

Hello,

Solving polynomial equations of degree higher than 2 (quadratic)
generally involves some guesswork. For example, you first try the
RATIONAL ROOT TEST to see if it has any rational roots. If none, then
the problem is certainly more difficult, and with luck, (or if the
problem is designed to be solvable), you factor it into a product of
QUADRATIC polynomials.

RATIONAL ROOT TEST

The only possible rational roots of a polynomial (set equal to zero)
are rational numbers whose:

(i) numerators are divisors of the CONSTANT term

(ii) denominators are divisors of the COEFFICIENT of the HIGHEST
degree term.

For the present case, since the constant term and the leading
coefficient are both 1, we need only test the rational numbers 1 and
-1. Direct calculation shows that neither of these satisfies the
equation, so the equation has NO rational root.

The only  way to solve the equation is FACTORIZATION into a product of
two QUADRATIC factors.  Generally, we would try:

(x^2 + ax + b)(x^2 + cx + d)

However, in the present case, observe that the polynomial is
SYMMETRIC. So, we try to see if it is possible to arrange these
quadratic factors to be SYMMETRIC as well. In other words, we try to
factor it in the form:

(x^2 + ax + 1)(x^2 + cx + 1)

Expanding this product, we have:

(x^2 + ax + 1)(x^2 + cx + 1)
= x^4 + ax^3 + x^2 + cx^3 + acx^2 + cx + x^2 + ax + 1
= x^4 + (a+c)x^3 + (ac + 2)x^2 + (a+c)x + 1.

Comparing with the given polynomial, we would like to have:

a + c = -6
ac + 2 = 11 ---> ac = 9

The only possibility is  a = c = -3.

If you can see this immediately, that is wonderful, and you should
proceed directly to the next paragraph.

If not, you find these numbers a and c by first eliminating one of
them, and see that you run into a QUADRATIC equation:

-6 - a = c
a(-6-a) = 9
-6a - a^2 = 9
a^2 + 6a + 9 = 0
(a+3)^2 = 0
a = -3

From this, c = -3 as well.

This means that the given polynomial is indeed a SQUARE:

x^4 - 6x^3 + 11x^2 - 6x + 1 = (x^2 - 3x + 1)^2

Its roots are therefore those of  x^2 - 3x + 1 = 0

Now, by the quadratic formula, we get:

x = (3 +/- Sqrt 5)/2.

Each of these is counted twice as a root of the 4th degree equation.

-Doctor Yiu,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials

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