Find Other Roots of a3-3a2+a+5=0 Given 2-i

[Seirsly]

Homework Statement

Given one root of the equation, find the others.
a³-3a²+a+5=0
Given root: 2-i

The Attempt at a Solution

I know the answers, which are 2+i and -1. And I understand how to get the 2+i since when you square root anything, the answer can be positive or negative, but I have no idea where the -1 came from, or how to really solve the problem.

 

[jgens]

Well, since this is a cubic equation, you know that it must have at least one real root. Since one root is complex, this means that it only has one real root. Assuming you know how the root 2 + i is arrived at (to be certain, please provide an explanation), the rest is pretty simple. We know that x = 2 - i and x = 2 + i are solutions to your original function and that if we can factor them out then we should be left with the real root. Do you know any way to do this?

 

[Feldoh]

Well for starters note: $$(x-2+i)(x-2-i)(x+1) = x^3-3x^2+x+5$$(Although you would not know to include the factor (x+1) had you not known -1 is a solution, but for the sake of the argument it makes is much easier to understand)

You are correct as to why 2+i is also a root. You know that the given polynomial is a cubic which means three roots. You have two so it's obvious that there is one more to find.

Also note that you know the third root of the equation has to be real since imaginary roots must come in conjugate pairs (a +/- bi)

So what if we solve the equality $$(x-2+i)(x-2-i)(x+1) = x^3-3x^2+x+5$$ for the third factor:

$$\frac{x^3-3x^2+x+5}{(x-2+i)(x-2-i)}$$ -- Seems reasonable right? We can solve this by using the long division that you learned all those years ago!

 

[Seirsly]

Unfortunately, no, I have no idea. Last algebra class (very smart education system, placing geometry between algebra classes) was about a year and a half ago.

2+i was just given in the question as one of the given roots. I have no idea how to arrive at that answer. Where did the a's go?

Edit:
Thank you Feldoh! I'm reading your post now and after I process this a bit more, I think I understand the solutions.

 

[jgens]

Okay, I'll provide an explanation. Given any polynomial with an odd degree, we know that that polynomial must have at least one real root. Now, suppose we were to factor this root out of the equation, in your case we should be left with a quadratic equation, correct? We also know that each root to a quadratic equation must satisfy x = [-b (+/-) sqrt(b² - 4ac)]/2a. The question tells us that one root must be 2 - i, do you understand how it follows that 2 + i is also a root given the last few facts?

To find the last root, you need to determine the quadratic equation the produced the roots 2 + i and 2 - i and factor it out of the cubic equation (do this by dividing the polynomials). Once you do that, you should be left with the remaining real root!

 

[Seirsly]

I'm sorry for asking, most likely, idiotic questions, but how would you factor out 2-i? Like Feldoh has done?

Oh my, this math has left my head spinning... Turns out I have no idea about this question at all.

 

[Feldoh]

Well "a" is just a variable right? Well so is "x". You "a" and my "x" mean exactly the same thing: a=x.

You're trying to solve the equality: $$a3-3a2+a+5=0$$ for the values of a which when inputted into the equation gives a solution of zero.

Once again let's assume you all ready know the answer: a can equal (2+i), (2-i), and -1

Now, by writing it in the form: $$(a-(2+i))(a-(2-i))(a+1)$$ it's completely equivalent to $$a^3-3a^2+a+5$$ (if you don't think it is expand it out!) which means that:

$$(a-2+i)(a-2-i)(a+1) = 0$$

Now here comes the part that trips a lot of people. I claim that if all three of the below equations are satisified:

(a-2+i) = 0
(a-2-i) = 0
(a+1) = 0

Then $$(a-2+i)(a-2-i)(a+1) = 0 = a^3-3a^2+a+5$$ Because if we solve any of those expressions for "a" we get:

a = 2+i
a = 2-i
a = -1

Which indeed are the answers!

Now to generalize this a little bit if a polynomial (just a fancy word for a function with a few a's raised to different powers, i.e. 5a^5+a^2 or a^2+2a+2, etc...)

is satisfied by some number b (b is just a variable, it stands for some number! In your case b = 2+i, 2-i, and -1)

such that a = b then (a-b) is a factor of the polynomial.

So back to your problem. Assume that we don't know that a=-1 is a solution. We still know that a=2 +/- i are solutions which means that we can represent:

$$a^3-3a^2+a+5$$

as

$$(a-2+i)(a-2-i)(another solution) = a^3-3a^2+a+5$$

So it makes sense that we just divide both sides by (a-2+i)(a-2-i) so we get:

$$(another solution) = \frac{a^3-3a^2+a+5}{(a-2+i)(a-2-i)}$$

 

[jgens]

Yes, you would do it just as Feldoh has done! To make it simpler, determine the quadratic equation that produced the roots (2 + i) and (2 - i).

 

[rock.freak667]

Right well, long division is one way, but here is a similar method which is quicker called synthetic division.

so one root is 2+i, so the conjugate of the complex number is another root. Hence 2-i is a root. So your cubic can be factored as three linear factors, (a-(2+i)) , (a- (2-i) ) and (Pa+Q) where P and Q are constants. (You want to know P and Q to find the other root).


Hence
a³-3a²+a+5 = (a-(2+i))(a-(2-i))(Pa+Q)

to make life simple instead of expanding out the entire right side. In the right side, the product of the first terms will give you the first term on the left. i.e. (a)(a)(Pa) = a³ => P=1.

The product of the last terms on the right side gives the last term on the left side.
i.e. -(2+i)*-(2-i)(Q) = 5, you can get Q and find the third root.

Another method you could have used is this:

For $Ax^3+Bx^2+Cx+D=0 \ with \ roots \ \alpha,\beta,\gamma$

(your equation is the same, just that I replaced 'a' with 'x')

the sum of the roots is given by -B/A

so that $\alpha+\beta+\gamma = \frac{-B}{A}$

You know what two of the roots are, so you want to find the third. You have the above equation relating the roots, so the third is easy to find.

 

[HallsofIvy]

Another way to do this is: knowing that 2- i is a root and all coefficients are real, it follows that 2+ i is also a root. That means that x- (2-i) and x-2+ i are factors of [itex]x^3- 3x^2+ x+ 5[itex]. (x-(2-i))(x-(2+i))= ((x-2)-i)((x-2)+i)= $(x-2)^2- (i^2)= x^2- 2+ 2$. Now divide $x^3- 3x^2+ x+ 5$ to find that the quotient is x+ 1= x-(-1).[/itex][/itex]

 

[deancodemo]

Here is an easier way.
Given 2-i is one root, then another root is 2+i (since coefficients are real, complex roots occur in conjuagate pairs). Now, call the last root x. Sum of roots = (2-i) + (2+i) + x = -b / a = 3. Upon simplifying, x = -1. Done!