# Tricky Factoring Questin (X^4 + 4)

1. Mar 12, 2012

### Plutonium88

now i found how to do the solution by doing an "Unorthodox" method of completing the square, could some one explain why the guy called it unorthodox?

Solution:

x^4 + 4

x^4 + 4 +4x^2 -4x^2 (completing the square)

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

(x^2 + 2)^2 - (2x)^2

(x^2 +2 - 2x)(X^2 + 2 + 2x)

Ans: (x^2 +2x - 2)(x^2 +2x +2)

Now initially i didn't get this question on my tests. So i'm curious, how would i have known or how can i change my way of thinking so that i could use this technique, because i still don't exactly get it completely.

In completing the square i remember

ax^2 + bx + c

so now you take

(b/2)^2 = a number

ax^2 + BX + C -(B/2)^2 + (b/2)^2

and then factor... in this question though, you have no b term... so yea can some one just point me in the direction i need to be thinking?

Last edited: Mar 12, 2012
2. Mar 12, 2012

### Dickfore

$$z^4 + 4 = 0$$
$$z^4 = -4 = 4 \, e^{i \pi}$$
$$z_1 = \sqrt[4]{4} \, e^{i \frac{\pi}{4}} = \sqrt{2} \, \left[ \cos \left( \frac{\pi}{4} \right) + i \sin \left( \frac{\pi}{4} \right) \right] = 1 + i$$
$$z_2 = \sqrt[4]{4} \, e^{i \frac{3 \, \pi}{4}} = -1 + i$$
$$z_3 = \sqrt[4]{4} \, e^{i \frac{5 \, \pi}{4}} = -1 - i$$
$$z_4 = \sqrt[4]{4} \, e^{i \frac{7 \, \pi}{4}} = 1 - i$$
$$z^4 + 4 = \left[ (z - 1)^2 + 1 \right] \, \left[ (z + 1)^2 + 1 \right]$$

3. Mar 12, 2012

### scurty

That's only true if a = 1, make sure you divide through first!

Think about what happens in this multiplication: $(a + b)^{2} = a^{2} + 2ab + b^{2}$. The middle term is simply $2 \cdot \sqrt{a^{2}} \cdot \sqrt{b^{2}}$. Does that help you see what you need to add? And depending on if you want $(a+b)^{2}$ or $(a-b)^{2}$, you would add in $+2ab$ or $-2ab$, respectively.

4. Mar 13, 2012

### Plutonium88

But A^2 + B^2 is not the same as (A+B)^2

like how can i verbally what is going on when i add a +4x^2 and a -4x^2. what is a good way to word the instructions of doing that..

5. Mar 13, 2012

### scurty

Exactly. $a^{2} + b^{2} = a^{2}+2ab+b^{2}-2ab = (a+b)^{2} - 2ab$.

In your examply in your first post you had $x^{4}+4=(x^{2})^{2}+(2)^{2}$. Can you now see why you add and subtracted $4x^{2}$? Whenever you have the sum of two squares, you know to add 0 creatively to compress the expression up.

6. Mar 13, 2012

### Plutonium88

ohh so its just a strategy to add something which relates to the equation, but whatever you add to the equation, must satisfy the equation by being equal to 0?

7. Mar 13, 2012

### Plutonium88

What type of factoring is this man? this loooks cool? something involving radians?? :O explain to me please?

8. Mar 13, 2012

### scurty

Yep! When you complete the square, you add a constant term so you can compress some terms into a sum that is squared. In this case you need to add the middle terms (being multiplied by x) so you can do the same thing.

9. Mar 13, 2012

### Plutonium88

Okay thanks a lot man i really appreciate the use of your time. Thank you.

10. Mar 13, 2012

### Dickfore

I am using the "fundamental theorem of algebra". Every polynomial of degree n has exactly n roots in the set of complex numbers. If you find them, then you could factorize it as:
$$P_n(z) = (z - z_1)(z - z_2) \ldots (z - z_n)$$
To find the roots in this simple case, you need to find the 4th root of -4. I used the exponential form to find the roots and Euler's formula after that.

Now, when the coefficients of the polynomial are real numbers, as is the case here, the roots come in complex conjugate pairs. You can combine the linear factors of the pair to make a quadratic factor:
$$(z - a - b i)(z - a + b i) = (z - a)^2 - (i b)^2 = (z - a)^2 + b^2$$