# Finding polynomial function's zero values

not native English speaker so I may get some terms wrong and so on.

On to the question:

I have as an assignment to find a polynomial function $f(x)$ with the coefficients $a$, $b$ and $c$ (which are all integers) which has one root at $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$.

I've done this with $f(x) = 0$ where $x = \sqrt{a} + \sqrt{b}$ through an iterative method which looks like this (forgive me for my pseudo code):

1. var x = sqrt(a) + sqrt(b)
2. Set x to x multiplied by the conjugate of x
3. Repeat from step 2 until all square roots are gone

The full calculation looks like this (the exponent signs disappeared, sorry about that):

$(x-(\sqrt{a}+\sqrt{b}))(x+(\sqrt{a}+\sqrt{b})) = x^2-(\sqrt{a} +\sqrt{b})^2 \\ = x^2-(a+2\sqrt{a}\sqrt{b} +b) \\ = x^2-a-b-2\sqrt{a}\sqrt{b}$

$((x^2-a-b)-(2\sqrt{a}\sqrt{b} ))(( x^2-a-b)+(2\sqrt{a}\sqrt{b} )) = x^4-2ax^2-2bx^2+2ab+a^2+b^2-4ab \\ = x^4-2ax^2-2bx^2-2ab+a^2+b^2 \\ = x^4-2x^2(a+b)-2ab+a^2+b^2$

$p(x) = x^4-2x^2(a+b)-2ab+a^2+b^2$

And then I used the factor theorem to calculate the remaining roots, which gave the following results: $x = (\sqrt{a} +\sqrt{b} )$, $x = -(\sqrt{a} +\sqrt{b} )$, $x = (\sqrt{a} -\sqrt{b} )$ and $x = -(\sqrt{a} - \sqrt{b})$.

When I try the same method on $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$, the calculations just become absurd. Any kind of help would be enormously appreciated!

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Subtract $\sqrt a$ from both sides, square both sides: now you have $\sqrt a$ on one side, $\sqrt{bc}$ on the other. Solve for $\sqrt a$, square both sides: now you have only $\sqrt{bc}$. Solve for $\sqrt{bc}$, square both sides, voila! all square roots gone.
If you need to know the other zeros, the full set is $\pm\sqrt a\pm\sqrt b\pm\sqrt c$ where the signs are to be taken independently of each other (making 8 zeros in all).
You certainly have the right idea. Following Gerry Myerson's suggestion above, we have \begin{align} x&=\sqrt{a}+\sqrt{b}+\sqrt{c}\\ x-\sqrt{a}&=\sqrt{b}+\sqrt{c}\\ (x-\sqrt{a})^2=x^2-2x\sqrt{a}+a&=b+2\sqrt{bc}+c=(\sqrt{b}+\sqrt{c})^2\\ x^2+a-b-c&=2x\sqrt{a}+2\sqrt{bc}\\ \left(x^2+a-b-c\right)^2&=4x^2a+8x\sqrt{abc}+4bc\\ \left(x^2-a+b+c\right)^2-4x^2a-4bc&=8x\sqrt{abc}\\ \left(\left(x^2-a+b+c\right)^2-4x^2a-4bc\right)^2&=64abc\cdot x^2. \end{align} Expanding this will yield a polynomial with integer coefficients (assuming $a$, $b$, and $c$ are integers) which has a zero at $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$.