Question

How do you write a polynomial with zeros: 2, 4 + sqrt5, 4-sqrt5?

Answer 1

`x^4-14x^3+67x^2-130x+88`

Explanation:

Note that: if `a` is a zero of a polynomial, then `(x-a)` is a root of the polynomial. Applying this to the zeros at hand, we see that that polynomial is equal to

`(x-2)(x-4)(x-(4+sqrt5))(x-(4-sqrt5))`

We can simplify the first two through distributing. We will get to the second two, which is the more challenging part of this problem, but first distribute the minus sign.

`=(x^2-6x+8)(x-4-sqrt5)(x-4+sqrt5)`

`=(x^2-6x+8)((x-4)-sqrt5)((x-4)+sqrt5)`

Notice that the last two terms are in the form `(a-b)(a+b)`, which equals `a^2-b^2`. Here, `a=x-4` and `b=sqrt5`. This gives us

`=(x^2-6x+8)((x-4)^2-5)`

`=(x^2-6x+8)(x^2-8x+16-5)`

`=(x^2-6x+8)(x^2-8x+11)`

Now, we can distribute these terms. There will be a lot, so hang on tight.

`=overbrace(x^4-6x^3+8x^2)^((x^2-6x+8) * x^2)+overbrace(-8x^3+48x^2-64x)^((x^2-6x+8) * -8x)+overbrace(11x^2-66x+88)^((x^2-6x+8) * 11)`

Combining all the like terms `x^3` with `x^3`, and so on, our result is

`=x^4-14x^3+67x^2-130x+88`

Note that this polynomial can be multiplied by any constant, and the roots will be the same. However, this is the simplest form of the polynomial.