How to create a polynomial with a variable root

Question

If we have $f(x) = a + bx + cx^{2}$ and $g(x) = m + nx + ox^{2}$ with roots $s$ and $t$ respectively, how does one make a polynomial with, say, the roots $(s+t)$ or another polynomial with $st$? What is the logic behind it?

Answer 1

If you want to have roots $r_1, r_2\dots r_n$ you can just write out the following:

$$ (r_1 - x)\times(r_2- x)\times\dots(r_n -x)=0$$

And multiply all that together.

Roots of an function $f(x)$ are the values of $x$ for which the function equals $0$.

For example, see that if you choose to set $f(x)=(x-1)(x-2)$, its roots are $1$ and $2$.

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For example, for an equation with roots $s$ and $t$, write $(x-s)(x-t)=0$ which yields:

$$ x^2-x(t+s)+st=0$$

You can replace the $0$ with $f(x)$ to graph it.

Answer 2

A root of a polynomial is a spot where the polynomial is equal to $0$. Thus, if we factor the polynomial, we will get a term which is $(x-s)$. So, we can write out a polynomial as the product of all these terms.

So, since the polynomial $$P(x)=c*(x-a_1)(x-a_2)(x-a_3)...(x-a_n)$$ will be $0$ at each of it's roots, the roots are simply $a_i$.

Applying this to your example where the roots are $(s+t)$ a polynomial which fits the description would be $$P(x)(x-(s+t))$$ Where $P(x)$ is any polynomial, as we will have a root at $(s+t)$ in addition to any of the roots of $P(x)$

Answer 3

If $f(x) = a + bx + cx^2$ and $g(x) = m + nx + ox^{2}$ has roots $s$ and $t$, then $f(x) = (x-s)h(x)$ for some $h(x)$ and $g(x) = (x-t)k(x)$ for some $k(x)$. And if you want a polynomial with a root $s+t$, then your polynomial should look like $(x-(s+t))v(x)$ for any $v(x)$.