Find polynomial of lowest degree

Question

find the polynomial of lowest degree having roots 3+i, and 3-i and 3 having lead cofficeint of 1.

Answer 1

We begin by taking factors of the polynomial, which are based on the roots.

The factors are (x-(3+i)), (x-(3-i)) and (x-3).

So the equation becomes:  y=(x-(3+1))(x-(3-i))(x-3).

We simplify:  y=(x-3-i)(x-3+i)(x-3)

I see a short cut here, using difference of two squares, and come up with the next simplification:

y=((x-3)^2-i^2)(x-3)

Since (x-3)^2=x^2-6x+9, and since i^2=-1, the equation looks like this:

y = (x^2-6x+9 - -1)(x-3) = (x^2-6x+9 +1)(x-3) = (x^2-6x+10)(x-3) = x^3 - 9x^2 + 28x -30

When you graph this, it crosses the x-axis at x=3, which is why one of the roots is x=3.  The other two crossings are "imaginary", so you won't see them on the graph.

Basically, the way to do these problems is to take whatever root there is, subtract it from x (x-r), and make that a factor in your polynomial, then multiply all the factors out until you get your polynomial equation.

Answer 2

Okay, you'll start by setting up the equation using the roots.

 

(x - (3+i))   (x - (3-i))  (x-3)

 

now, multiply out those irrational roots and cancel the matching terms (marked with ** and xx)

 

(x-3-i)(x-3+i)

= x(x-3+i) + -3(x - 3 + i) + -i(x - 3 + i)

= x^2  + -3x  + ix +  -3x + 9 + -3i + -ix + 3i + -(-1)

                        **                    xx     **    xx             

= x^2   + -3x + -3x + 9 + 1

= x^2 -6x + 10

 

That's the first two factors, now multiply in the third factor

 

(x^2 -6x + 10 )(x-3) = x(x^2 -6x + 10 ) + -3(x^2 -6x + 10 )

= x^3 -6x^2 +10x + -3x^2 + 18x -30

= x^3  -9x^2 +28x -30

 

And check your conditions - coefficient of the lead term is invisible, so it must be 1.  Check!

 

Now, technically, at this point, you'd substitute in each of the roots and verify you have the right answer... but I'll only do that for the root 3 right now.

 

27 -9(9) +28(3) -30

27 - 81 +84 -30 = 0  check.