Question

How do you write a polynomial with zeros -6, 3, 5 and leading coefficient 1?

Answer 1

`x^3-2x^2-33x+90`

Explanation:

This has `3` roots, so it will be a polynomial of degree `3`. If we factor a 3rd degree polynomial we get:

`(x+a)(x+b)(x+c)` where a , b and c are constants:

We can find these constants using the known roots. If the roots are:

`alpha=-6`

`beta=3`

`gamma=5`

Since the constant term in each factor will have an opposite sign to the root, we have:

`(x-alpha)(x-beta)(x-gamma)`

Substituting `alpha , beta and gamma` for root values we have:

`(x-(-6))(x-(3))(x-(5))=(x+6)(x-3)(x-5)`

The coefficient of the first term is the product of the coefficients of the x terms in each factor, since these are all `1`, the coefficient of the first term will be `1`

`(x+6)(x-3)(x-5)=x^3-2x^2-33x+90`