Find the values of the zeros of a polynomial

Question

QUESTION: Two of the zeros of x⁴ - 12x³ + 59x² - 138x + 130 are a+ib and a+2ib. Find a and b and hence, factorise the polynomial over the real field. 

 

I assume that you just need to make P(a+ib) and P(a+2ib) equal to zero to find the values, but this proved to be quite long and complicated. Also, it is implied by the imaginary number 'i' that the roots are complex, so how can the polynomial be factorised over the real field?

Answer 1

Since we already have our zeros, we can use them by setting them as a product that gives us 0.  The form in which the zeros are written in indicate that they were obtained using the quadratic formula.  Because of this, we have conjugates.  The conjugates will give us the last two zeros.

 

(x - [a + ib])(x - [a - ib])(x - [a + 2ib])(x - [a - 2ib]) = 0

 

 

Next, we expand the left side using FOIL.  Beware that we also have terms that are difference in perfect squares.  That will make it easier to expand.

 

[x² - x(a + ib) - x(a - bi) + (a² + b²)] * [x² - x(a - 2ib) - x(a + 2ib) + (a² + 4b²)] = 0

 

 

Now we  simplify the left side.

 

(x² - 2ax + a² + b²)(x² - 2ax + a² + 4b²) = 0

 

 

Note that we no longer have any i terms in the expansion.  Next, we expand further so that we have polynomial with a degree 4.  The best way to do this is to rewrite the left side using distributive property.

 

x²(x² - 2ax + a² + 4b²) - 2ax(x² - 2ax + a² + 4b²) + a²(x² - 2ax + a² + 4b²) + b²(x² - 2ax + a² + 4b²) = 0

 

x⁴ - 2ax³ + a²x² + 4b²x² - 2ax³ + 4a²x² - 2a³x - 8ab²x + a²x² - 2a³x + a⁴ + 4a²b² + b²x² - 2ab²x + a²b² + 4b³ = 0

 

 

Next, combine like terms and set this equal to the original function.  But first, we group all the x⁴, x³, x², x, and constant terms together.  We do this so that we can equate coefficients.  Equating coefficients will allow us to solve for a and b.

 

x⁴ + x³(-2a - 2a) + x²(a² + 4b² + 4a² + a² + b²) + x(2a³ - 8ab² - 2a³ - 2ab²) + (a⁴ + 5a²b² + 4b³) = 0

 

x⁴ - 4ax³ + (6a² + 5b²)x² + (-4a³ - 10ab²)x + (a⁴ + 5a²b² + 4b³) = x⁴ - 12x³ + 59x² - 138x + 130

 

 

Now we can equate coefficients.

 

x³  ------->     -4a = -12

x²   ------->     6a² + 5b² = 59

x   ------->     -4a³ - 10ab² = -138

const ---->       a⁴ + 5a²b² + 4b³ = 130

 

You have 4 equations in the system to work with.  Use them to solve for a and b.  Once you have a and b, plug them into this factored form of the equation.

 

(x² - 2ax + a² + b²)(x² - 2ax + a² + 4b²) = 0

 

 

I know its lengthy, but follow these steps and you should get your actual factorization.

 

Answer 2

I thought long and hard about whether to post an alternative solution when Michael did an extremely lengthy but nevertheless correct solution. In the end, I figure that the student can appreciate that there are different ways to solve a problem.

 

First, note that since the polynomial is a quartic, there are a total of four roots.

Second, since two of the given roots are imaginary and unequal, the other two are the complex conjugate of these roots.

Hence the roots are

a + ib, a - ib, a + 2ib, a - 2ib

Take the first two.

Sum = 2a       Product = a² + b²

The sum and product form the quadratic expression  x² - 2ax + a² + b²

 

Take the next two.

Sum = 2a       Product = a² + 4b²
The sum and product form the quadratic expression x² - 2ax + a² + 4b²

 

The required polynomial is

P(x) = (x² - 2ax + a² + b²)(x² - 2ax + a² + 4b²)

 

Multiplying out

P(x) = x4 - 4ax³ + (6a² + 5b²)x² - (4a³ +10ab²)x + a⁴ + 5a²b² + 4b⁴

 

Compare the coefficients of x's in this equation with the coefficients of the x's from the original polynomial

P(x) = x4 - 12x³ + 59x² - 138x + 130

For instance, the coefficients of x³

x³:   -4a = -12                    a = 3

The coefficients of x²

x²: 6a² +5b² = 59

      6(3)² + 5b² = 59

                  5b² = 5             b =1 

 

Answer 3

 

To answer your second question, may I suggest you graph the given equation (y = x⁴ - 12x³ + 59x² - 138x + 130) on a graphing calculator or at an online graphing website such as https://www.desmos.com/calculator.  Look at the line and ask yourself, do you see any point where it touches the x-axis (this is the definition of a zero, or root).  Secondly, I suggest you ponder on the definition of an imaginary number, as being something imaginary which of necessity does NOT exist, namely, the square root of negative one (√(-1)).  Can something that does not exist be used to explain reality?  Which is more outrageous, using imaginary numbers in real mathematics, or the lunatic in the asylum who somberly affirms that they are the President of the United States, and acts out the role with unflinching conviction throughout the course of their life, making major and minor decisions from this premise; and, incidentally also happens to claim that the sun is inhabited by penguins not one whit different from our native earth species?  In conclusion, the above polynomial cannot be factored over the real field or have any real usefulness in anyone's life, anymore than imaginary numbers can exist nor earth-based penguins live in the sun.

 

Instead of making up imaginary numbers, whoever was the mad mathematician who started this, it would have been a much greater service to humanity for them to have explored the beautiful implications of the fact that there exist polynomial expressions which implicitly exclude the possibility of finding a root because they can, intrinsically, never be equal to zero.  

 

For instance, to find the real solution to the problem I would start with the equation:

 

x⁴ - 12x³ + 59x² - 138x + 130 ≠ 0

 

Then:

 

x⁴ - 12x³ + 59x² - 138x ≠ -130

 

From the graph we can also quickly see that:

 

x⁴ - 12x³ + 59x² - 138x + 130 ≥ 4

 

and:

 

x⁴ - 12x³ + 59x² - 138x ≥ -126

 

 

What is the whole point of finding roots, or zeros of a polynomial?  Isn't it the goal of finding a solution to the problem, expressed as x = something, or when there is not one solution at least narrowing it down to the only possible set of solutions?  Is it the philosophical implications of "there is no solution" that frightened the mathematician into creating an "imaginary" solution?  Probably not.  But compare the graph of this equation to the graph of:

 

 x⁴ - 12x³ + 59x² - 138x + 126

 

This equation has one root, which is 3.  So it can be solved for x if P(x) =0.  When you look at it it is an exact copy of the first equation just moved down 4.  So to get a solution to the first equation, all we would have to do is temporarily move the equation four down on the y axis.  Isn't that easier, in the end, than doing countless mind numbing mental gymnastics with things that don't exist?  If I wanted to move the equation down some more I could also get it to give me two real roots.  In the end, this approach makes more sense, because it tells me all the useful information I could possibly want to know about my polynomial, whereas chasing around imaginary numbers mostly results in headaches and school drop outs.