# Factor this

Discussion in 'Math' started by Ratch, Jun 28, 2008.

1. ### Ratch Thread Starter New Member

Mar 20, 2007
1,068
4
To the Ineffable All,

The math forum has not moved for a long time. This problem has stumped a lot of people. How about you? Ratch

Factor x^4 + 64

2. ### mik3 Senior Member

Feb 4, 2008
4,846
69
do you mean factorize it?

3. ### Ratch Thread Starter New Member

Mar 20, 2007
1,068
4
mik3,

Yes. You are right. I should have use the verbal form of the word instead of the noun. Ratch

4. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Factor is also a verb according to my dictionary (American Heritage Dictionary of the English Language).

Meanwhile, I am looking to see if I can factor x$^{4}$ + 64.

5. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
x$^{4}$x + 64
= (x - (2 + 2i)) (x - (2 - 2i)) (x - (-2 + 2i)) (x - (-2 - 2i))

Obviously, the factors have coefficients that are not real numbers, but the problem statement placed no restrictions on the factors.

The complex numbers in the inner parentheses are the complex fourth roots of -64, which arise from finding the zeroes of the equation x$^{4}$ + 64 = 0, or x$^{4}$ = -64. These roots all have magnitude 2$\sqrt{2}$ and have angles of, respectively, 45 degrees, 135 degrees, 225 degrees, and 315 degrees.

6. ### Ratch Thread Starter New Member

Mar 20, 2007
1,068
4
Mark44,

Yes, your answer is correct, and I should have specified that complex conjugate terms were not necessary. It factors easily into two quadratic expressions. Can you tell me how one could solve it? Ratch

7. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
How's this? (x$^{2}$ + 8i) (x$^{2}$ - 8i)
The two quadratic factors can be further factored into the linear factors in my earlier post.
Mark

8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
523
I think Ratch was looking for

(x^2-4x+8)(x^2+4x+8)

Which do not factorise further in the real world.

9. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Studiot, one should be able to multiply selected pairs of the linear factors I gave to come up with the two quadratics you showed. "One" here means I haven't done this.

Nov 9, 2007
5,005
523
11. ### Ratch Thread Starter New Member

Mar 20, 2007
1,068
4
Studiot did what I was looking for. I was not looking to find the roots, only to factorize the expression. I don't know what the method of undetermined coefficients is, but my method is called completing the square. This problem came from a high school textbook about 50 years ago.

x^4+64 = x^4+16x^2+64 -16x^2= (x^2+8)^2-(4x)^2

Which is the difference of two squares, and easily expands to studiot's answer. Ratch

12. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
If so, you should have stated that you wanted factors with real coefficients.

Finding the roots of a polynomial and factoring it are nearly identical problems. For a polynomial an*x$^{n}$ + a(n-1)*x$^{n-1}$ + a(n-2)*x$^{n-2}$ + ... + a1*x + a0, if the roots are r1, r2, r3, ..., rn, then the factored form is (x - r1)*(x - r2)*(x - r3)*...*(x - rn).

If you know the roots of a polynomial, you know its factorization.

Some of you might be wondering why the polynomial Ratch gave has so many factorizations. I provided two of them, and Studiot provided another. The linear factorization I provided earlier in this thread is irreducible. Each of the other two factorizations (the ones with quadratic expressions) can be factored further into linear factors, as long as you allow coefficients from the complex numbers.

If we're working with integers instead of polynomials, there is a similar situation. 100 can be factored in several ways:
4 * 25
5 * 20
2 * 50
and a few more.

There is one factorization that is better than all others:
2 * 2 * 5 * 5
Except for the order in which the factors appear, this factorization into primes is unique. If factors that aren't primes are allowed, then a given factorization is not guaranteed to be unique.
Mark

13. ### Ratch Thread Starter New Member

Mar 20, 2007
1,068
4
Mark44,

Yes, I should have specified factorizing into two quadratics. Ratch