Question

How do you construct polynomial equations with the given roots?

1. `2`, `4` and `-7`.
2. `5` and `sqrt(3)`.

Answer 1

1. `x^3+x^2-34x+56 = 0`

2. `x^3-5x^2-3x+15 = 0`

Explanation:

Note that if a polynomial in `x` has a zero `a` then it has a factor `(x-a)` and vice versa.

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For question 1 we can construct a polynomial:

`f(x) = (x-2)(x-4)(x+7) = x^3+x^2-34x+56`

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this `f(x)`.

So the polynomial equation:

`x^3+x^2-34x+56 = 0`

has roots `2`, `4` and `-7`.

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For question 2 we can multiply out `(x-5)(x-sqrt(3))` but this will result in a polynomial with irrational coefficients:

`(x-5)(x-sqrt(3)) = x^2-(5+sqrt(3))x+5sqrt(3)`

If - as is probably the case - we want a polynomial with integer coefficients, then we also need the rational conjugate `-sqrt(3)` to be a zero and `(x+sqrt(3))` a factor.

Then we can define:

`g(x) = (x-5)(x-sqrt(3))(x+sqrt(3)) = (x-5)(x^2-3) = x^3-5x^2-3x+15`

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this `g(x)`.

So the polynomial equation:

`x^3-5x^2-3x+15 = 0`

has roots `5`, `sqrt(3)` and `-sqrt(3)`.