Thread: Problems concerning polynomials / complex numbers

1. Problems concerning polynomials / complex numbers

f(x) = |x+5| + |x+2| + |x| + |x-3|

The equation gives us the total length of a cable between the points A and B

P=-5
Q= -2
O=0
R= 3

x is the centre of the cable.

The question is, where should the centre of the cable be to minimize toe total length of cable to all four points. And: what is the minimum (of the length of the cable)

Also, how can I graph this function that has absolute values in its equation?

Last part of this problem: A point S is added 7km away from O (so it adds to the equation + |x-7|. Where should it be for a minimal cable length?

Next question:
I need to find a fourth degree polynomial with the rational coefficients 2- i*squareroot 3 and squareroot2 +1 as two of the four zeros. How do I do this and what is the solution? Please help me.

2. Hello, Instigator!

Here's #2 . . .

2) Find a fourth-degree polynomial with the rational coefficients
. . where $2 - i\sqrt{3}$ and $1 + \sqrt{2}$ are two of the four zeros.

You're expected to know these facts . . .

If $2 - i\sqrt{3}$ is a zero, then its conjugate, $2 + i\sqrt{3}$, is also a zero.

If $1 + \sqrt{2}$ is a zero, then its conjugate, $1 - \sqrt{2}$, is also a zero.

The polynomial is: . $P(x) \;=\;\left[x - (2-i\sqrt{3})\right]\cdot\left[x - (2+i\sqrt{3})\right]\cdot\left[x-(1 + \sqrt{2})\right]\cdot\left[x - (1-\sqrt{2})\right]$

Multiply and simplify: . $\boxed{P(x) \;\;=\;\;x^4 - 6x^3 + 14x^2 + 10x - 7}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If the thought of that multiplication fills you with Fear and Loathing,
. . let me show you a trick for handling "conjugate multiplication".

The first product is: . $\left[x - (2 - i\sqrt{3})\right]\cdot \left[x - (2 + i\sqrt{3})\right]$

. . Regroup it like this: . $\left[(x-2) + i\sqrt{3}\right]\cdot\left[(x-2) - i\sqrt{3}\right]$ . . . This is: $(a+b)(a-b)$

. . So we have: . $(x-2)^2 - (i\sqrt{3})^2 \;\;=\;\;x^2 - 4x + 4 -(-3) \;\;=\;\;{\color{blue}x^2-4x + 7}$

The second product is: . $\left[x - (1 + \sqrt{2})\right]\cdot\left[x - (1 - \sqrt{2})\right]$

. . Regroup it like this: . $\left[(x-1) - \sqrt{2}\right]\cdot\left[(x-1) + \sqrt{2}\right]$ . . . This is: $a^2-b^2$

. . So we have: . $(x-1)^2 - (\sqrt{2})^2 \;\;=\;\;x^2-2x+1-2 \;\;=\;\;{\color{blue}x^2-2x-1}$

Then we can multiply: . $\bf{{\color{blue}(x^2-4x+7)(x^2-2x-1)}}$ . . . see?

3. Thank you very much, especially for the trick and the great explanation!