Question 373504
The standard form for complex numbers is a+bi. A complex number in the form a-bi is called to complex conjugate of a+bi.<br>
There are two important facts needed for all three of your problems are:<ol><li>If a zero of a polynomial with real coefficients is complex, then the complex conjugate is also a zero.</li><li>Let q = a zero of a polynomial with real coefficients. Then (x-q) is a factor of that polynomial.</li></ol>Let's see how this works on your problems:
1. 3i and square root of 3
3i, in standard form is 0+3i. Since this is a zero and we are looking for a polynomial with real coefficients, then 0-3i, or -3i, is also a zero. So we have three zeros: 3i, -3i and {{{sqrt(3)}}}. So (x-3i), (x-(-3i)) and (x-{{{sqrt3)}}} are factors of the polynomial. And if we want the polynomial of lowest degree, these are the <i>only</i> factors of the polynomial. So:
{{{P(x) = (x-3i)(x+3i)(x-sqrt(3))}}}
For standard form we need to multiply this out. The first two factors fit the pattern: {{{(a+b)(a-b) = a^2 - b^2}}}. So I will take advantage of this to multiply them quickly:
{{{P(x) = ((x)^2-(3i)^2)(x-sqrt(3))}}}
which simplifies as follows:
{{{P(x) = (x^2-9i^2)(x-sqrt(3))}}}
Since {{{i^2 = -1}}}:
{{{P(x) = (x^2-9*(-1))(x-sqrt(3))}}}
{{{P(x) = (x^2+9)(x-sqrt(3))}}}
To multiply what is left, we use FOIL:
{{{P(x) = x^2*x + x^2*sqrt(3) + 9x - 9*sqrt(3))}}}
or
{{{P(x) = x^2*x + sqrt(3)*x^2 + 9x + (-9*sqrt(3))}}}<br>
2. 5,-1, and -2+2i
Since -2+2i is a zero, then -2-2i is also a zero. With these four zeros, the polynomial of least degree is:
{{{P(x) = (x-5)(x-(-2))(x-(-2+2i))(x-(-2-2i)))}}}
or
{{{P(x) = (x-5)(x+2)(x-(-2+2i))(x-(-2-2i)))}}}
Using the pattern again with the factors for the complex roots we get:
{{{P(x) = (x-5)(x+2)(x^2-(-2+2i)^2)}}}
or
{{{P(x) = (x-5)(x+2)(x+2-2i))(x+2+2i)}}}
Multiplying the last two factors is not as hard as it looks if you look at it like {{{(red(x+2)-2i))(red(x+2)+2i). Looking it the red parts as "a" and the 2i as "b", this fits the (a+b)(a-b) pattern! SO we know that it will multiply out to {{{a^2-b^2}}}:
{{{P(x) = (x-5)(x+2)((x+2)^2-(2i)^2)}}}
which simplifies as follows:
{{{P(x) = (x-5)(x+2)((x^2+4x+4)-4i^2)}}}
{{{P(x) = (x-5)(x+2)((x^2+4x+4)-4(-1))}}}
{{{P(x) = (x-5)(x+2)(x^2+4x+4+4)}}}
{{{P(x) = (x-5)(x+2)(x^2+4x+8)}}}
Multiplying the first two factors, using FOIL, we get:
{{{P(x) = (x^2-3x-10)(x^2+4x+8)}}}
I'll leave the rest of the multiplying for you. Just multiply each of the three terms of the first factor by each of the three terms of the second factor. (Yes, that is 9 multiplications!) Then add any like terms.<br>
Using the given zero, find all other zeros of f(x)
3. i is a zero of f(x)= x^4-4x^3+2x^2-4x+1
If i is a zero then so will -i. So we have two factors. This 4th degree polynomial should have 4 zeros so we have two more zeros to find. To find these other zeros, we will factor P(x).<br> We will be using synthetic division to factor P(x). We will divide by the two known factors: (x-i) and (x+(-i)):
<pre>
i |  1  -4       2      -4      1
---      i      -1-4i    4+i   -1
    -----------------------------
     1  -4+i     1-4i    i      0
</pre>
So at this point {{{P(x) = (x-i)(x^3 + (-4+i)x^2 + (1-4i)x + i)}}}. We will divide the second factor by the second zero:
<pre>
-i |   1  -4+i     1-4i    i
----        -i       4i   -i
     ------------------------
       1  -4       1       0
</pre>
Now {{{P(x) = (x-i)(x-(-i))(x^2-4x+1)}}}
or
{{{P(x) = (x-i)(x+i)(x^2-4x+1)}}}
The third factor will not factor further (at least not easily). But it is a quadratic so we can use the Quadratic Formula:
{{{x = (-(-4) +- sqrt((-4)^2-4(1)(1)))/2(1)]]]
which simplifies as follows:
{{{x = (-(-4) +- sqrt(16-4(1)(1)))/2(1)]]]
{{{x = (-(-4) +- sqrt(16-4))/2(1)]]]
{{{x = (-(-4) +- sqrt(12))/2(1)]]]
{{{x = (4 +- sqrt(12))/2]]]
{{{x = (4 +- sqrt(4*3))/2]]]
{{{x = (4 +- sqrt(4)*sqrt(3))/2]]]
{{{x = (4 +- 2*sqrt(3))/2]]]
{{{x = (2(2 +- sqrt(3)))/2]]]
{{{x = (cross(2)(2 +- sqrt(3)))/cross(2)]]]
{{{x = 2 +- sqrt(3)}}}
So the remaining zeros are {{{2 + sqrt(3)}}} and {{{2 - sqrt(3)}}}<br>
FWIW, with these last two zeros we can finish factoring P(x):
{{{P(x) = (x-i)(x+i)(x-(2+sqrt(3)))(x-(2-sqrt(3)))}}}
or
{{{P(x) = (x-i)(x+i)(x-2-sqrt(3))(x-2+sqrt(3))}}}