Factoring Polynomials with Complex Roots?

Question

I have no idea what in the world I'm doing.
I've missed math twice during the lecture on this, so I'm confused as to what I'm supposed to be looking for, since the book barely gives an explanation.

Somehow this: x^3-x^2+x-1 = (x-1)(x-i)(x+i)

I really want to know the whole process as to how to go about this. I've looked up and down this website already, and I see everything about roots. Would I need to find the roots to factor these?

Here are two problems from my homework that I must factor:

x^2+x+2

x^3+6x^2+11x+6

I don't even know where to start. :(
Someone please help me and guide me through this process. I have a test over this at the end of the week, and I really want to know how to do this.

Thank you soo much. :]
...If someone helps me. :(

Answer 1

It is common to say that polynomials with irrational or complex roots are prime. Which doesn't mean that they can't be factored; it simply means that the factors are irrational. Easiest is to deal with quadratics; you can always equate the expression to 0, and use the quadratic formula to get the roots: if they are R1 and R2, then the factors are (x-R1)(x-R2). Your first expression, x^3-x^2+x-1 has a root at x = 1, so (x-1) is a factor and the other factor is (x^2+1). Now the roots of the quadratic are of course +i and -i, giving the factors which you noted. The second expression can be solved directly by the quadratic formula, the roots are complex and involve sqrt(7). The third one has a root at x = -1; use synthetic division to pull (x+1) out of the expression and use the quadratic formula to whack what is left. In this case, the roots are real integers.

Answer 2

Factoring Complex Polynomials

Answer 3

Assuming you know the general solution to a quadratic equation, y = [-b +/- sqrt(b^2-4*a*c)]/(2a)
cases will arise when b^2-4*a*c<0. In this case, you just don't toss the book away and scream, you solve and use the term "i" to refer to sqrt(-1).

As with any other quadratic, if the FACTORS are
(AX+B) and (CX+D), the ROOTS are X= -B/A and
X=-D/C. What you see above are ROOTS.


With higher polynomials, like the cubic you cite, you also can factor the equation and come up with imaginary terms such as you see. There IS a general cubic solution, but it's a real dog to work with. Expanding from quadratic eqs, the roots are 1, i and -i. There is a theorm that says that complex roots come in pairs, so if you have one imaginary root, you must have two.

The quadratic doesn't factor easily, so you should use the formula to get the roots.

The cubic looks promising, assume that one factor is (x+2). Now divide the cubic by (x+2) to get a quadratic, which hopefully has nice factors.

Answer 4

If one of your root is -3, then your polynomical could be rearranged to be the product of two polynomials, (x+3) * (a degree 2 polynomial) (The "x+3" is where your -3 root comes into play). Then you divide your polynomial x^3 -x^2 -7x + 15 by the (x+3) bit and obtain the parabolic that can be solved like any quadratic.

Answer 5

x^3-x^2+x-1

we already know that (x-1) is a root
--------x^2+1
________________
(x-1)| x^3-x^2+x-1
-------x^3 -x^2
---------------0+x-1
------------------x-1
------------------0

so now you have
(x-1)(x^2+1)

now need to factorise x^2+1=0

x=sqrt-1
x=+/- i

so wirting it all out
(x-1)(x-i)(x+i)



x^2+x+2

you can just use quadratic formula

(-1+/- sqrt(1^2-4*1*2))/2*1

-1/2+/-1/2sqrt(-7)
-1/2+/-(sqrt(7)/2)*i
so then put x=-1/2+/-(sqrt(7)/2)*i
giving (x+1/2+(sqrt(7)/2)*i)(x+1/2 - (sqrt(7)/2)*i)=0


polynomial division for the last one gives all real answers
(x+1)(x+2)(x+3)

Answer 6

i just always do the trial and error method,.... i really don't know that too