Do mean find the polynomial given its roots ? If so the answer is
(x -r1)(x-r2)...(x-rn) where r1,r2,.. rn is the given list roots.
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The "roots" of a polynomial are the solutions of the equation polynomial = 0. That is, any value which you can replace for "x", to make the polynomial equal to zero.
You forgot to copy the polynomial. However, the Fundamental Theorem of Algebra states that every polynomial has at least one root, if complex roots are allowed. If a polynomial has only real coefficients, and it it of odd degree, it will also have at least one real solution.
Rational roots
4
The real roots of what, exactly? If you mean a square trinomial, then: If the discriminant is positive, the polynomial has two real roots. If the discriminant is zero, the polynomial has one (double) real root. If the discriminant is negative, the polynomial has two complex roots (and of course no real roots). The discriminant is the term under the square root in the quadratic equation, in other words, b2 - 4ac.