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Nonlinear Factors

Date: 10/12/98 at 15:28:45
From: Adam Turner
Subject: Factoring (x to the fourth + 4)

I have been told that factoring the sum of two "squared" numbers is 
not possible; however, my instructor indicates this one can be done.  
I have had no luck. Is he pulling my leg?

Date: 10/13/98 at 02:00:50
From: Doctor Pete
Subject: Re: Factoring (x to the fourth + 4)


First, I'll write:

     x^4 + 4

to mean, "x raised to the fourth power, plus 4."  (Similarly, x^3 
means "x cubed," and x^2 means "x squared.")

Now, usually, when we think of factoring a polynomial, we are thinking 
of finding a form:

     (x - a)(x - b)(x - c)...

where a, b, c, ... are constants. But factoring is closely connected 
with a similar problem, which is finding roots of a function. In 
particular, if you have a polynomial function f(x), and solve the 

     f(x) = 0

then what you are finding are the values of a, b, c,... in the factored 
form. For example, say:

     f(x) = x^2 - 4

Then, observe that f(2) = f(-2) = 0; that is, x = 2, and x = -2 are 
roots of f(x). Then f(x) has the factored form:

     f(x) = x^2 - 4 = (x - 2)(x + 2)

However, in the case where:

     f(x) = x^4 + 4

we see that there is no real value of x for which f(x) = 0, because x^4 
is always nonnegative. So one cannot expect a factorization of the 

     (x - a)(x - b)(x - c)(x - d)

where a, b, c, d are all real numbers. However, it may be possible to 
factor it as:

     (x^2 + px+ q)(x^2 + rx + s)

that is, as a pair of irreducible quadratics. Why this may be possible 
will be clearer in a moment. For now, suppose that:

     x^4 + 4 = (x^2 + px+ q)(x^2 + rx + s)

for some unknown constants p, q, r, s, which are all real numbers. By 
expanding the righthand side, we see that:

     x^4 + 4 = x^4 + (p+r)x^3 + (q+s+pr)x^2 + (ps+qr)x + qs

If we equate the coefficients on both sides of this equation (why?), 
then we find that:

          p + r = 0
     q + s + pr = 0
        ps + qr = 0
             qs = 4

From the first equation p = -r, and substituting this into the third 
equation gives r(-s + q) = 0. This means r = 0, and/or q = s. 

Suppose r = 0. Then p = 0, and we have from the second equation that 
q + s = 0. But this is impossible, since qs = 4. So we must have 
instead that q = s, and hence, from the fourth equation, q = s = 2 or 
q = s = -2. 

Suppose q = s = -2. Then from the second equation, we have that 
-2 - 2 + pr = 0, which implies pr = 4. Since r = -p, we find that  
-p^2 = 4, which is impossible. 

Now suppose, p = r = 2. From the second equation, we then have 
2 + 2 + pr = 0, or pr = -4.  Since p = -r, it follows that r^2 = 4. 
Therefore, r = 2 or -2, and p = -2 or 2. Note that this makes sense, 
since this produces the factorization:

     x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)

and clearly p and r are interchangeable, as long as they have opposite 
signs. One can check that the factorization is correct by multiplying 
out the righthand side.

Now, to gain an understanding of why the polynomial x^4 + 4 factors 
into two quadratics but not four linear factors, we attempt to find the 
roots of, say, x^2 + 2x + 2, which is one of the quadratics. Clearly, 
any root of this polynomial is also a root of x^4 + 4. Using the 
quadratic formula, we find that the roots are given by:

     x = (-2 + Sqrt[-4])/2 , (-2 - Sqrt[-4])/2
     x = -1 + I , -1 - I

where I is the square root of -1. These numbers are complex numbers, 
which explains why a linear factorization can't be found. Now, one 
could look at the original quartic:

     x^4 + 4

and solve for the roots directly: let w = x^2, so we find:

     x^4 + 4 = w^2 + 4

Immediately we see that w = {2I, -2I}, and hence x = {Sqrt[2I], 
-Sqrt[2I], Sqrt[-2I], -Sqrt[-2I]}. Without going into too much detail, 
these values are simply:

     x = {-1 + I, -1 - I, 1 + I, 1 - I}

But these complex numbers come in complex conjugate pairs, and when 
multiplying and adding complex conjugates, they become real numbers.  
So, when we multiply the two linear factors:

     (x - (-1 + I))(x - (-1 - I))

for example, we obtain x^2 + 2x + 2, an "irreducible" quadratic.  

I hope that made some sense.

- Doctor Pete, The Math Forum
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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