1. ## Polynomials and x+i

I could use some help with this:

Find a polynomial of degree three with real number coefficients and zeros of 3 + i and -7.
Kind of having trouble with the real number coefficients, and how to find them?
(x+7)
(x - (3+1))
(x + (3-1))
Is that how you write 3 + i as a zero?

Any help is greatly appreciated! Thanks!

2. Originally Posted by Såxon
I could use some help with this:

Find a polynomial of degree three with real number coefficients and zeros of 3 + i and -7.
Kind of having trouble with the real number coefficients, and how to find them?
(x+7)
(x - (3+1))
(x + (3-1))
Is that how you write 3 + i as a zero?
Not quite, you write it like this:
$(x + 7)[x - (3 + i)][x - (3 - i)]$.

There is a formula regarding multiplying two binomials that contain complex conjugates:
$[x - (a + bi)][x - (a - bi)] = x^2 - 2ax + (a^2 + b^2)$

So
$[x - (3 + i)][x - (3 - i)] = x^2 - 6x + 10$.

Multiply that by x + 7 to get
$(x + 7)(x^2 - 6x + 10) = x^3 + 7x^2 - 6x^2 - 42x + 10x + 70 = x^3 + x^2 - 32x + 70$.

01

3. Originally Posted by Såxon
I could use some help with this:

Find a polynomial of degree three with real number coefficients and zeros of 3 + i and -7.
Kind of having trouble with the real number coefficients, and how to find them?
(x+7)
(x - (3+1))
(x + (3-1))
Is that how you write 3 + i as a zero?

Any help is greatly appreciated! Thanks!

For $x = -7$ to be a root, then $x + 7$ is a factor.

For $3 + i$ to be a root, then $x - (3 + i)$ is a factor.

Since complex solutions always appear as complex conjugates, that means $3 - i$ is also a solution. So $x - (3 - i)$ is another factor.

So your polynomial is of the form

$0 = a(x + 7)[x - (3 + i)][x - (3 - i)]$

Expanding gives

$0 = a(x + 7)[x^2 - (3 - i)x - (3 + i)x + (3 + i)(3 - i)]$

$0 = a(x + 7)(x^2 - 6x + 4)$

$0 = a(x^3 - 6x^2 + 4x + 7x^2 - 42x + 28)$

$0 = a(x^3 + x^2 - 38x + 28)$

This equation will hold true for any value of a.

For simplicity, let $a = 1$ and then you have your cubic.

4. If $x_1=3+i$ is a root of the polynomial then $x_2=3-i$ is also a root.

Then $P(x)=(x-3-i)(x-3+i)(x+7)=x^3+x^2-32x+70$