Forming a polynomial from its roots

If \(f(x)\) is a cubic and the roots of \(f(x)=0\) are \(r_1\), \(r_2\) and \(r_3\), then \(f(x)=a(x-r_1)(x-r_2)(x-r_3)\), where \(a\) is the coefficient of \(x^3\).

Why: by the factor theorem, \(f(r)=0\) means \(x-r\) is a factor. Three such factors make a cubic, so only the leading constant \(a\) is left to find.

Method

1. Write \(f(x)=a(x-r_1)(x-r_2)(x-r_3)\), using a letter for any unknown root. For a quartic with two given roots and a given quadratic factor such as \(x^2+1\), keep that factor as it is: \(f(x)=a(x-r_1)(x-r_2)(x^2+1)\).
2. Use the given leading coefficient for \(a\), or use a given value or remainder (the remainder theorem) to form an equation.
3. Solve for the unknowns and expand if the question asks for descending powers.
4. Check one given condition with your final \(f(x)\).

Example

The roots of the cubic equation \(f(x)=0\) are \(-1\), \(2\) and \(3\), and \(f(0)=12\). Find \(f(x)\).

\(f(x)=a(x+1)(x-2)(x-3)\); \(f(0)=a(1)(-2)(-3)=6a=12\), so \(a=2\).

\(f(x)=2(x+1)(x-2)(x-3)=2x^3-8x^2+2x+12\). Check: \(f(0)=12\).